Classical Mechanics - ut.edu.sa · The Classical Mechanics Labs is a laboratory, experimental, course designed to accompany the ... rotations, oscillations and waves) and will be

Classical Mechanics A Lab Manual

Fall 2011

A. Abdesselam

Spring 2012

Cover Photo: http://www.npr.org/blogs/krulwich/2011/05/16/136342435/cinderellas-ball-this-time-with-pendulums . see the beautiful demonstration(s) there.

Copyright © Abdelouahab Abdesselam ([email protected])

Table of Contents Introduction ............................................................................................................ 1

Measurement and Error Analysis ............................................................................ 3

Measurement and Error Analysis (Experiment I) .................................................. 19

Measurement and Error Analysis (Experiment II) ................................................. 21

Free Fall Acceleration ........................................................................................... 22

Understanding Motion ......................................................................................... 26

Newton’s Second Law ........................................................................................... 31

Viscosity Coefficient of a Liquid ............................................................................ 35

Conservation of Mechanical Energy ...................................................................... 40

Hooke’s Law and Harmonic Motion ...................................................................... 46

Simple Pendulum .................................................................................................. 51

Static Equilibrium .................................................................................................. 55

Sound waves ......................................................................................................... 60

Specific Heat Capacity of a Solid ........................................................................... 68

Mechanical Equivalent of Heat ............................................................................. 72

References ............................................................................................................ 77

Appendix A: How to write a report ....................................................................... 78

Appendix B: How does a Vernier caliper work? .................................................... 89

Appendix C: Least Square Fit ................................................................................. 92

Appendix D: Simple Pendulum .............................................................................. 99

1

Introduction

The Classical Mechanics Labs is a laboratory, experimental, course designed to accompany the theoretical lectures given in the Classical Mechanical Theory course. It therefore presents the student with unique opportunities to have a firm grasp of the abstract concepts and laws he/she is exposed to during the lectures. In this course indeed the student experiments with static equilibrium, different types of motion (constant speed motion, uniformly accelerated motion, rotations, oscillations and waves) and will be briefly introduced to thermodynamics. These subjects are nowadays considered very basic but nevertheless essential to every serious sciences student. They should constitute an important part of the basement for further high and more involved studies. The laboratory practices are important for another reason. It is the occasion where the students can learn precious experimental techniques. They will develop measurement and data analysis skills. It is not rare in physics that the measurement of a quantity (distance, speed, acceleration,…etc.) put us in front of insurmountable challenges. Only very highly qualified people, genius at times, can imagine, design and build apparatus that make such measurements realizable. The Data analysis and the extraction of useful physical quantities could also be arts in their own right. By doing and contemplating lab experiments the students acquire some of these skills which are also required in dealing with advanced studies and researches. Sciences and Physics in particular would not have made the enormous progresses apparent everywhere around us without uncountable number of meticulous experiments giving sometimes unpredicted results. On another side, it often happens that many mathematical theories present themselves as candidates to describe the physical reality. But it is only by doing experiments that we know which theory is acceptable and which are not. The experiment has thus the privilege of saying the last word. The student should be highly motivated and organized to be able to benefit from this course. One way to demonstrate a sense of organization is to keep an log book (note book) for the physics labs. This is the place where student record the data, make sketches, express his/her thoughts etc. It should be hand be written and needs not to be clean but should be clear though. The aim of log book is to contain enough information such that the student could at any later time reconstruct the experiment. Another benefit is that the log book contains data which are not easily obtainable elsewhere. Although this course is a basic one, the equipments hereby used cannot be found in anyone’s kitchen.

2

Finally the students should after each experiment return a well written report which is graded by the instructor. These grades contributes to from 30% to 50% of the final grade. Therefore it is of the students’ interest to take the reports quite seriously.

3

Measurement and Error Analysis

This is a quick introduction to Measurement and Error analysis. The material presented here is

crucial for conducting the experiments in the labs and correctly reporting their results. The

student should have a first superficial reading of the content and then come back whenever a

more accurate understanding is required. There are many books1 dealing in detail with the

subject of “Data and Error Analysis” which the serious student should consult to deepen her/his

knowledge. The student is also encouraged to look for other resources on the internet.

1 How do we Report a Measurement The Data taking or measurement process is an essential step in conducting an experiment and

should be given a great deal of attention and care. Even so, all measurements do have errors

attached to them because of the inherent incapability of any instrument to give the true value

exactly (of course the aim of any measurement is to evaluate the true value). A measurement is

therefore incomplete, to say the least, if it is not accompanied by an estimation of the

committed error. An acceptable measurement of a quantity should then consist of the

measured value, sometimes also called best value, and its error :

2 The Meaning of Error What do we mean by ? We can first give a rough answer and say that we

mean we do have some confidence that the true value of the quantity is somewhere within

the interval [ ]. This is illustrated in Figure 1.

Figure 1: the true could be anywhere within the uncertainty interval

1 See References at the end of this manual.

4

Said otherwise, this means that if we repeat the experiment many times and calculate , we expect

that the majority of the results (of ) will lie within the mentioned interval. A more precise

definition will be given later in this document (see the paragraph entitled “The statistical interpretation

of the error”).

The error above is given in an absolute form, i.e. without comparison to the measured value .

It’s sometime more informing to quote the committed error as a percentage or relative error

(evaluated as

). This, for example, makes it easier to judge if the measurement is a

good or poor one. A relative error of few per cent (i.e. less than 10%) is acceptable in our labs.

On the contrary an error of more than 15% indicates a poor measurement. It’s also mandatory

to use the relative errors when comparing the precisions of the measurement of dimensionally

different quantities. In other words, to be able to say which of such quantities is measured with

a better precision, one has to look at their respective relative errors.

3 Why is it Important to Know the Error There are at least two reasons why we should know the value of the committed error.

1. The first reason is that the error gives us an idea on the quality of a measurement. A

measurement with a relatively big error is almost useless. Only a precise

measurement, i.e. one with a relatively small error, could be used with benefit in

practice. As said previously, in general, a relative error less than 10% can be

considered as acceptable in our labs.

2. The second reason is that we need the errors whenever we want to say something on

the equality of two measurements. It’s because we know the errors that we can make

a qualitative statement about the two measurements in figure 2-a and say that they

are probably equal. Similarly, because we know the value of the errors, we can

confidently say that those in figure 2-b are most likely to be different. A more precise

criterion in comparing two measurements is discussed in the paragraph entitled

“Comparing two measurement” (see below).

Figure 2: using the errors in comparing two measurements

5

Now, if the errors are that important, how should we proceed to estimate them? Here we

distinguish two situations: Estimating the error on directly measured quantities and indirectly

measured quantities. We treat each in turn.

4 How do we Estimate the Error on a Directly Measured Quantity We have just seen that it’s very important to know the value of the error committed in any

measurement. We therefore proceed now to discuss how we can evaluate such an error. There

are three different ways of estimating the error on directly measured quantities depending on

the instruments or methods we do use to take the measurements:

4.1 Analog Devices: Examples of these are the ruler, balance, Vernier caliper…etc. It’s, basically, any device that has

graduations on it. In our labs we will often take the smallest graduation as error. (See Giancoli,

4th Edition, page 3). This is because we cannot read, using the device’s scale, any value smaller

than the smallest graduation (there are no ticks to help us to do that). However, if you think

that this is a too big error, then consider taking half the graduation as error (or even a smaller

fraction). On the other side if you deem that your error should be greater than one graduation,

then take few graduations as error.

Example: if you are measuring the length of an object with sharp ends then a graduation as

error is generally safe. If, in contrary, the ends of the object are not sharp (i.e. they are not

clearly located) then it’s safer to take a few graduation as error.

4.2 Digital Devices Examples are the stop watch, multi-meter…etc. It’s any device that displays the measurement

as digits. Here one can take the smallest digit that can be displayed as error. For example, if the

instrument displays 1233, then one should report the measurement as 1233 . (this is

because the smallest number that this instrument is able to show is 0001 (=1)).

We can motivate this rule in the following way. Because the device in the example above

cannot display decimal numbers (it gave use the number 1233 without a decimal point), even if

1233 has a decimal part, it could not be shown on the device. So, the measured number 1233

could be in reality any one of the list 1233.0, 1233.1, 1233.2, 1233.3, 1233.4, 1233.5, 1233.6,

1233.7, 1233.8, 1233.9. The device when it returns the number 1233, it does in fact some sort

of rounding to the real number (which could be any one of the above list, but we do not know

which one, because the device, through its limited precision, has lost the decimal part). Now we

can estimate the committed error by saying that it is of the order of the difference between the

highest and lowest value of the considered number: . And

that is why we write . (We could probably do a better estimation by taking an error

6

half of this and write the result as . But in our labs we prefer to stick to the simpler

rule of quoting an error equal to one full unit of the smallest displayed digit).

Note: it should be stressed here that error estimated by the previous prescriptions is often the

smallest error one should quote. It’s the error from the measuring instrument itself. One should

consider error from other sources (e.g. Human errors, time of reaction, difficulties to locate

precisely the ends of an object…etc.). To be more specific an example is given: when measuring

the time interval of an action (a moving object crossing two point), one should note quote the

error only from the stopwatch, which is often very small, but should take account for the

human time of reaction (reflex) and ability to synchronize events.

4.3 The Standard Deviation method. This method can, and maybe should, be applied whenever one does have at hand many

measurements (say ). (For example one has repeated the same measurement times). The

rational of this method is that the spread of the measurements themselves should contain the

information on the precision of the measuring instrument. We have, therefore, just to find a

way to extract the error from a set of measurement. In fact, there already exists a well known

method, Standard Deviation method, which proceeds in three steps as follow:

a. First one calculates the mean. This will be your best estimate of the quantity under measurement2:

(1)

b. Then one calculates the standard deviation. This tells you how spread your measurements are (It

represents an estimate of standard deviation of the parent distribution3 and thus the error of one

measurement).

2 This can be shown by using the method of Maximum Likelihood. We assume, as is often the case, that the

measurements are distributed following a Gaussian distribution with means and standard deviations (for instance both unknown). With this assumption, the probability for making one observed measurement is

√

. This follows from the statistical interpretation of the measurements and their

errors (with the Gaussian distribution as parent distribution). The probability for making all the observed ’s is, following the probability multiplication rule, the product of the individual probabilities

∏

√ ,

-

,∑

- ∏

√

. The best estimate of the parameter is

the one that gives the maximum probability . To find this maximum we set the derivative with respect to to

zero:

,∑

- ∏

√

. This gives

∑

or ∑

. Solving this

equation we get

∑

. (Here we abuse the symbols and use the same symbol for the true and estimated

value of the mean).

3 A similar calculation as in note 1 above shows that the best estimator of is given by √

∑

.

(Here too we abuse the symbols and use the same symbol for the true and estimated value of the standard deviation).

7

√

(2)

Why do we use and not just in the definition of is explained in detail in one of the reference

books4.

c. Finally one calculates the error on the mean5.

√ (3)

The result can then be written as . Please note that Eq. 3 shows that the more

measurement you do (this is ) the smaller your error will be ( decreases as √ .

Note: If it happens that all of the measurements give the same value and thus is null, one

should be careful to note quote a zero error. In this case you should use method one of the

previous two methods). One should also use the previous methods whenever the standard

deviation method gives a smaller result (i.e. one should be on the safe side and report the

biggest obtainable error).

4.4 Number without error If we are given a number without the error being specified, we can then assume that the error

is equal to a unit of the least significant figure of that number (significant figures are described

a bit further down in this introduction). Here are a few examples:

1. 45678 has an error 1 .

2. 53.3 has error 0.1

3. 0.7830 has error 0.0001

4. 3450 has error 10

We can justify this rule in the following way (similar to the digital device rule explained above).

Consider the number given in item 1 above, i.e. 45678. Whoever gave us this number, he is not

sure about the digit that comes after 8. It could be any number between .0 and .9. he is not

sure about it and that’s why he did not specify it. Therefore 45678 could be in fact 45678.0 or

45678.1 or 45678.2,…, or 45678.9. This defines then the uncertainty in 45678. We can say that

4 See Appendix E in “An Introduction to Error Analysis: The Study of Uncertainties in Physical Measurements”, J. R.

Taylor, 2nd

ed. Briefly, this is needed to make the estimator give exactly the true value of the distribution

parameter when an infinite number of experiment is used (

). 5 Eq.3, which gives the error on the mean, can be proved in the following way. The error on

∑

comes

from the errors on the single (i.e. . We can then use the method of propagation of errors (see Eq. 4)

√∑

, where . We have

(

∑

)

. Then √∑

√

√

8

the error is of the order of the difference between the highest and lowest value of the

considered number: . And that is why we write

.

5 Statistical interpretation of the error This is the right place to continue the discussion we have started in the paragraph entitled “The

meaning of the error” (see above). A more precise definition of the error relies on the assumption6 that

the distribution of the results of repeated measurements of a quantity follows a Gaussian (or Normal)

curve. Within this assumption, we expect that 68.2% of the results of the measurements lie in the

interval [ ], 95.4% within the bigger interval [

], and 99.6% inside the even bigger interval [ ], where and

are the mean and the standard deviation of the distribution, respectively7. This statistical

interpretation of measurement is depicted in figure 3.

Figure 3:Gaussian distribution of repeated measurements

The error on the mean (or best value) can howevr be much less than and is given by

√ . (The error get smaller with increased number of measurements . See the previous

section the validity of which is in fact based on the assumption that the measurements are

distributed following a Gauss distribution). The values of , obtained in repeated

experiments8, also follows a Gaussian distribution, but with smaller standard deviation (to wit

√ ). This is shown graphically in figure 4.

6 This is indeed the case for many practical situations.

7 We abuse the language and neglect the difference between the (parent) distribution parameters and their

estimators. 8 By experiment we mean a set of measurements (like the ones discussed in figure 2). Each experiment gives us a

different value of . By repeating the experiment many times we get a distribution similar to what is shown in figure 3.

9

Figure 4: statistical interpretation of the error

Please note the differences with respect to the previous figure in both vertical and horizontal

axes labels. Figure 3 deals with measurements within one and only one experiment (where

each experiment consists of many measurements). In each experiment we estimate .

Figure 4, on the other side, deals with many of such experiments. It shows the distribution of

the ensemble of obtained in the many experiments. The Gaussian distribution tells us

that we should expect that 68.2% of the results of the repeated experiments will lie in the

interval [ ], 95.4% within the bigger interval [

], and 99.6% inside the even bigger interval [ ].

6 How do we Estimate the Error on an indirectly measured

quantity This is the case whenever the quantity in question, which is not measured directly, depends on

other quantities (or variables) which we can measure directly, i.e., they are of the type treated

above. The error on this functional quantity (that is a function of other quantities) is estimated

by propagating the errors of the directly measured variables to the functional quantity.

Assume that is a function of the independent variables . This is symbolically written

as . This means that the value of is readily calculated from the measured value

of It can be then shown9 that the error on is given by10:

9 The error on the quantity is

√ √

∑

where . We can expand the

function about the point in Taylor series:

we then get √∑ *

+

√

∑ [ (

) (

) (

)(

) ]

. If

the variables are independent, as we have assumed all along this document, then terms like

∑ * (

) (

)+

give, on the average, zero (because we would have as many positive

contributions as negative ones). We thus find √(

)

(

)

, which is Eq. 4.

10

√

(4)

where

denotes the derivative of with respect to while maintaining all the

other variables constant (it is called the partial derivative of with respect to . Equation (4)

says that knowing the errors on the error on can be easily calculated. Let’s look at

few examples:

1.

where is a constant (i.e. depends neither on nor on ). This is the simplest form of . We have here

only one variable and the derivative is quite simple:

. Substituting this result in Eq. 4 we

get:

√

or

Example: we measure the radius of a circle to be . What is the circumference of the circle

and its error?

Answer: The circumference is . The error on is, using ,

. thus .

2.

where and are constants. We have now two variables and and the partial differential with respect to

is

(remember, when differentiating with respect to , we

treat the other variables as constant).

In a similar way

The substitution of these derivatives in (Eq. 4) gives

√ (5)

Example: An object is composed of two parts of mass and . What is the mass

of the compound object and its error?

Answer: The mass of the compound object is . The error on is,

applying Eq. 5, given by √

3.

10

Here and mean the same thing (i.e. error on )

11

where is constant (i.e. is proportional to the product of two variables and ).

Then

and

. Replacing these derivatives in Eq.

4 we get:

√ .

We can put this equation in an easy to remember form. We divide by and get:

√

√

(6)

Example: We have a rectangular surface of length and width .

What is the area of this surface and its error?

Answer: The required area is given by . The error on is, applying Eq. 6,

given by √

√

1.8 So we have

.

4.

where and are constants. Note that now depends on only one variable.

Then

. Replacing in Eq. 4 and dividing by , as done in the previous example, one gets

√

Or

(7)

Example: If we measure the radius of a disk to be (1.000 ± 0.001)m, what is the error on the area of the disk?

Answer: First the area is . The error is given by

√

√

, ie,

5.

Then

(

)

(

)

(

)

. As before, substituting in Eq.4 and

dividing by we get (check it!):

√

(8)

12

Note that this formula is similar to that of case 2 (see Eq. 6).

Example: A cube of volume of (27.00 ± 0.05) and mass (250.0 ± 0.1)g. What is the error on its density

?

Answer:

√(

)

(

)

√(

)

(

)

. Therefore .

6.

where and are constants. The derivative with respect to is

. Substituting

in Eq.4 we get:

√

Dividing by we find:

√

then

Note: if , we can write this as . We then can apply the formula above with

replaced with .

7.

where and are constants.

. Replacing in Eq.4 and dividing by we get:

√

Then

8.

where and are constants.

13

and thus, using Eq.4, we get

Similarly one can show that if then

9. One can continue in a similar way with other kind of functions.

7 Significant Figures Significant figures in a given number are the figures that we know reliably, i.e. we are not

(totally) uncertain about them. Take for example the following result of some measurement:

. The error tells us that we are quite sure about the first three figures

( and a bit less sure about the fifth one (that is 4). But we are totally unsure about what

comes on the right side of 4; it could be any digit between 0 and 9. So we have in all 4

significant figures: . This means that even if the result was given to us in another form

, we should round it and keep only 4 figures (

), because the error tells us there really are only 4 significant figures.

7.1 Counting significant figures The significant figure on the far left is the most significant one. The significant figure on the far right is

the least significant figure. All the digits between the most and least significant digits, including zeros,

are significant. The zeros on the left or right of a number may be problematic and we need clear rules to

remove any ambiguity in counting significant figures. We have essentially two cases:

1. Zeros on the left are not significant:

Example: 012 and 0.12 both have two significant figures. 0.001 has only significant figure (the

zeros on the left do not count!).

The reason for this rule is quite obvious for non-decimal number. We can discard the zeros

without changing the value or the precision of the considered number. For example 012 is the

same as 12 and thus both numbers should be considered as having two significant figures.

The rational for a decimal number (like 0.001) is more subtle. The zeros are not considered

significant because one can change the unit (i.e. use a smaller unit) and recast the number in a

non-decimal form (example: and this 0.001 has

only one significant figure). In that case the zeros just inform us about the used scale. (Do not

confuse this with errors. We are assuming that 0.001 expresses the result of a measurement and

not the value of whatever error).

14

2. Zeros on the right, usually, are not significant for a non-decimal number but they are significant

for a decimal number:

Examples: 1200 has 2 significant figure (by the first part of this rule). 0.10 has also two significant

numbers (the second part of the rule).

The rational of the first part of this rule is as follow: assume you are given the number 1234 and

are asked to round it to two significant figures (because of some error, we are not sure about the

last two figures 3 and 4). What would you do? You cannot of course write 1234 as 12, since this

totally changes the value of the number, but you can write it as 1200 and state that the zeros on

the right are not significant. Thus 1200 would have only two significant figures. The zeros in this

case are just place holders (telling us where are the position of units, tens, hundreds, etc.). they

are not the real number zero (like in 0, 1, 2, 3,…etc.).

Note: it may occur that a zero on the right (of a non-decimal) number is significant (i.e. it is really

the number zero and the zero place-holder). In this case a decimal point should be added at the

right end of the number (if the zero in 120 is significant then write it as 120.).

The situation is different for a decimal number (like 0.10). A zero, if it is there, is significant. If it is

not significant we should not write it in the first place (write 0.1 instead of 0.10). Therefore zeros

on the right of a decimal number are significant (you cannot add or drop them just like that! They

represent precision and precision costs money!).

Examples:

1. 15734 has 5 significant figures

2. 71.090 has 5 significant figures

3. 0.0051 has 2 significant figures



7.2 Operations with significant figures One has to be careful to not gain precision just by doing arithmetic operations. For example if one does

the operation 25/12 using a calculator one gets on the screen 2.0833333. One should not just copy this

number as it is given by the calculator, because it pretends far more precision than contained in the

original numbers 25 and 12. Instead one must round the result to express the original precision and

report in this specific case 2.1. When operating with numbers, a distinction is made between addition

one side and multiplication or division on the other side. In the latter case the number of significant

figures of the least precise factor is of interest while in the former case it is the position of the least

significant digit which is of importance. The following makes things clearer.

1. Multiplication and Division: when multiplying or dividing two numbers the result should

retain a number of digit equals to the number of significant figures of the least precise factor. This

rule can be understood by considering error propagation as follow: take for example the operation 8 × 8.

Since we are not given an error on both factors (8 and 8) we assume an error equal to a unit of the least

15

significant figure i.e. 1 (we then have 1 and ). The error on the product is

√(

)

(

)

(see error propagation above). This means that . On the other

side, following the actual rounding rule, we can directly write . This makes sense because

the number 60 means an error equal to 10 which is of the same order of the one we just calculated using

error propagation. In other words, is the same results as . Other rules, e.g. the rule for

addition, can be motivated using the same reasoning.

Examples11:

8 × 8 = 60 both factors have one significant figure, so should do the result (60 has only one

significant figure).

8 × 8.0 = 60 the first factor (8) has one significant figure and the second factor (8.0) contains two

significant figures. The result should, as stated above, retain only one significant figure.

8.015 × 8.0 = 64 the first factor (8.015) has four significant figures and the second factor (8.0)

contains two significant figures. The result should retain two significant figures.

8 / 2.0 = 4 the first factor (8) has one significant figure and the second factor (2.0) contains two

significant figures. The result should retain one significant figures.

8.6012/2.0 = 4.3 the first factor (8.6012) has five significant figures and the second factor (2.0)

contains two significant figures. The result should retain two significant figures.

2. Addition: when adding (or subtracting) two numbers the result should be rounded to

the position of the least significant digit of the least precise number.

Examples12:

1. 1 + 1.1 = 2 1 is significant to the ones place, 1.1 is significant to the tenths place. Of the two, the least

accurate is the ones place. The answer cannot have any significant figures past the ones place.

2. 1.0 + 1.1 = 2.1 1.0 and 1.1 are significant to the tenths place, so the answer will also have a number in

the tenths place.

3. 100 + 110 = 200 100 is significant to the hundreds place, while 110 is significant to the tens place.

Therefore, the answer must be rounded to the nearest hundred.

4. 100. + 110. = 210. 100. and 110. are both significant to the ones place (as indicated by the decimal), so

the answer is also significant to the ones place.

Note: From the previous examples one sees that one could lose precision when operating with two

numbers with different precision (number of significant figures). That is why it’s important to measure

the involved quantities to the same precision.

11

These examples are borrowed from Wikipedia (http://en.wikipedia.org/wiki/Significance_arithmetic) 12

These examples are copied from Wikipedia (http://en.wikipedia.org/wiki/Significance_arithmetic). More examples are listed there.

16

8 Types of Error, Precision and Accuracy We distinguish two types of error: random and systematic. We also define the physical meaning

of the precision and accuracy.

8.1 Random errors This is the type of errors we described its estimation in the previous sections. This kind of error

causes the result of repeated measurements to be randomly13 distributed around a mean value

(usually standing for the true value). Its origin lies in the inherent limitation of any apparatus to

produce the same value in all measurements. As we have seen above, see Eq. 3, the magnitude

of this error can be reduced by repeating the measurement many times.

8.2 Systematic errors This type of errors tends to shift systematically the mean value of the measurements towards

either smaller or larger values than the true value. In other words, the results of repeated

measurement will be all smaller than the true value or all larger. This means than we cannot

reduce this error by repeating the measurement. Usually a systematic error can be detected14

by analyzing carefully the experiment and by comparing the actual result with other similar

experiments. Once detected and estimated, the systematic error should be corrected for in the

measurements by either adding or subtracting its value to or from the best estimate. If the

actual error causes the measurements to be larger than the true value, then we should subtract

it, otherwise add it.

8.3 Precision and accuracy Precision and accuracy have different meaning in physics although they are interchangeable in

everyday language uses. The precision is related to the random error: The smaller this error is

the more precise the measurement will be.

The accuracy, on the other side, describes how far is the measured value from the true value.

The accuracy is rather related to the systematic error. Figure 1 sketches the difference

between precision and accuracy: the center of circles represents the true value and the dots

represent repeated measurements.

13

It often happens that the distribution is Gaussian. 14

Systematic errors are more difficult to detect than random errors.

17

Figure 5: accuracy versus precision

9 Comparing Two Measurements15 If two quantities and are exactly known (i.e. no errors is attached to them) then they are

equal if their difference is equal to zero:

(9)

But if it happens that we know the two quantities only approximately, i.e. an error is attached

to each of them, then the difference alone will not tell us if they are equal or not. For example,

the two quantities may be equal in reality but the measured values could produce a non-zero

difference. Think of two pieces of wood which do have the same length but, because of the

finite precision of any measurement, they are found to have different length. The equation

above gives then a non-null difference.

We need therefore an alternative way to test if two quantities are equal. We obviously have to

make use of the errors which we denote and . The error on is given by (see example 2

in the error propagation section):

√ (10)

We could then say that the measured quantities and are equal if their difference is within

the error

| √

15

Comparing two measurements can be a very serious and difficult business. Here we propose a rather naïve and intuitive method.

18

Or to have mathematically well behaved quantities

This can be written as:

(11)

where , pronounced chi-squared, is a random variable in the sense that if we repeat the

measurement of and and calculate every time, then obtained values will be randomly

distributed and follow a chi-square distribution. The mean value of the distribution is equal to

the number of degree of freedom for the situation at hand. In our case, Eq. 11, we have one

degree of freedom. Because of statistical considerations, we will say that two measurements

and are equal if and different otherwise i.e. we use 3 and not 1 as in Eq. 11 as follow:

The value 3 is chosen because if really equals , then by doing several measurement on

them, only less than 5% of the measurements will produce a greater than 3. Thus if it

happens that your is less 3, then you are almost sure that and are equal. (To improve

our confidence in the comparison outcome we need to either do more measurements and/or

improve a measurement precision).

Remark: If the prescription just given suggests to you that two values are different, but which

by other means you know are equal, then you may want first to make sure that you have not

underestimated your errors and second look for any possible systematic error that you may

have ignored in your measurements.

10 Fitting Procedure Fitting data means finding the theoretical curve which describes the data. Most of the fitting

you will do in this lab, if not all, are linear. This means that you assume that your data (or

measurements) are represented by a straight line. The fitting operation provides you with the

parameters of this line, namely, the slope and intercept. You can do that very easily by using

Excel. The theory behind the linear fitting, used by Excel, is described in Appendix B. you are

encouraged to have a look at it.

19

Measurement and Errors Analysis I

Pre-Experiment (1)

1 Aim of the experiment The aim of this session is to introduce you to significant error and error estimations.

2 Introduction The material needed for this lab has been explained in detail in the introductory chapter

“Measurement and Error analysis”.

3 Significant figures and errors 1. How many significant figure does each of the following numbers have:

a. 13748 …………

b. 71.090 …………

c. 0.0051 …………

d. 00063 …………

2. Round the previous numbers to at most three significant figures.

a. …………….

b. ……………..

c. ……………..

d. ……………..

3. Write properly the results of the following operations:

a. 456 + 38938 =

b. 0.4565 + 12.4 =

c. 34 x 124 =

d. 2346 / 233 =

e. 123.5 / 94 =

4 Error estimation 1. Measure the length of your logbook and estimate the error on it and write your final

result.

20

2. A series of measurements of a physical quantity gave us the following numbers (in some

unit):

17.47468 17.47392 17.47616 17.47499 17.47257

a. Calculate the best estimate of the physical quantity (i.e. calculate the mean value of

the series of measurements).

b. Calculate, using the standard deviation method, the error on the considered

quantity.

c. Write the error keeping only one significant figure. (It’s may be safest to always

round up the error).

d. Write the final result (i.e. best estimate absolute error).

21

Measurement and Errors Analysis II

Pre-Experiment (2)

1 Aim of the experiment The aim of this session is to introduce you to error and error propagation.

2 Introduction The material needed for this lab has been explained in detail in the introductory chapter

“Measurement and Error analysis”.

3 Error Propagation In general, if is a function of and : , the error on is given by

√

Applying the above formula, what is the error on dues to the errors on and in each of the following cases? You should write the derivation steps. (Hint: see “Measurement and Error analysis” chapter pages 6-9).

1. , and are constant. 2. , is constant.

3.

, is constant.

4 Application of Error Propagation 1. Measure the dimensions of the parallelepiped given to you (length , width and

height ) and estimate their error.

2. Calculate the volume of the parallelepiped and its error

3. Measure the mass of the parallelepiped and estimate its error.

4. Calculate the density of the parallelepiped

and its error .

22

Free Fall Acceleration

Experiment (1)

1 Aim of the experiment You measure in this experiment the free fall acceleration ( ) due to the gravity of the earth.

2 Introduction An object acted on by the earth gravity alone is said to be in free fall motion16. The attraction

force due to earth gravity is nearly constant near the earth surface17. This means, following

Newton’s second law, that the motion of falling object is uniformly accelerated. The Kinematics

of such motion is simple (see Giancoli, Chap. 2). Because the acceleration, say , is constant,

the speed of the object increases linearly with time18:

(1)

Where is the speed of the object at The distance covered by the object in a time is

obtained by integrating Eq. 119:

16

Examples are a thrown rock, the moon,…etc. 17

The gravity force between the earth and an object in its surrounding is

, where is a universal

constant, the earth mass, the object mas and is the distance between the center of the earth and the center of the object. If the object is near the earth surface at a hight then we can write , where R is the

earth radius (i.e. constant). We can write

(

), where we have used

a Taylor expansion. Now, since (i.e.

) we can neglect the terms that contains

beside 1, and thus we

can write

, which is constant. On another side Newton’s second law says that . Applying

this to the case at hand, where we have the only force of gravity, we get

and then

, the

gravity accelerarion. 18

The linear acceleration is the rate of change in the linear speed with respect to time:

. This means that

. Integrating both sides between time 0 and time we get where is the speed at time

. (the acceleration is constant so we can take it out of the integral sign: ∫

∫

gives

). 19

The linear speed is the rate of change in the linear displacement with respect to time:

. This means that

. This equation can be integrated as follow: ∫

∫

, is not constant because we assumed that

there exists acceleration, so it cannot be taken out of the integral sign. But we know, from the previous note, that

23

(2)

To use this equation in our experiment, we need to write it in a linear form. We can do so in the

following way:

Starting from Eq. 2, we subtract from both sides to get:

We then divide both sides by and get:

(3)

This equation is of the form which is the equation of straight line with slope and

intercept This means that if we plot

against (i.e.

vs ), we get a straight line20 from

which we can get the slope and thus the acceleration (Note that from the above equations

.

3 Equipment needed Two photo-light sensors connected to digital timer, holder, and small ball.

4 Experimental procedure

. Replacing this in the integral we get: ∫

∫

∫

∫

where of course

and are constant and can be taken out of the integral sign. This is easily integrated to give

which is the equation given in the text.

20 After the fitting procedure.

24

Figure 1: free fall setup

1. The setup should be already connected as shown in Fig.1. If this is not the case talk to the instructor.

2. Check that the digital timer is working fine by passing your hand through the first sensor (top one), to start the time (that is t=0), and then through the second sensor (bottom one), to stop the timer. The time elapsed between starting and stopping the timer will be displayed on the timer screen.

3. Set a distance of 20 cm between the two sensors (the distance should be measured, say, from the middle of one sensor to the middle of the other sensor). Do not forget to estimate your measurement error. (move the bottom sensor, the top one should stay fixed) .

4. Release the ball from a fixed distance, say 5 cm, above the top sensor. This distance should not be changing during the experiment. (why?). The ball should pass through the two sensors so that it starts and stops the timer.

5. Record the used distance and the corresponding time displayed on the timer in the table given below. Repeat steps 4 and 5 three times.

6. Repeat steps 3-5 for the distances shown in the table below.

Distance between

the two sensors

Time Average

Time

20

30

40

50

60

5 Data Analysis 1. Estimate the errors on .

25

2. Calculate, using the standard deviation method, the error on . (use the row that

would give you the biggest error).

3. Calculate the error on

.

4. Plot, using Excel,

versus time (i.e. on the X-axis, and

along the Y-axis) and

draw the error bars (on the same graph). Do not forget to put the titles and units on

the axes.

5. Fit linearly your data and get the slope and the intercept.

6. Using the Excel function linest, calculate the errors on the slope and intercept.

7. Does your data represent the theory given by Eq.3? Look at the fit quality variable .

8. Plot your data, error bars, fit line on a graph paper this time.

7. Calculate the acceleration and its error.

8. How does your result compare to the known value of .

6 Further questions

1. What is the value of the speed of the ball when it crosses the first sensor (hint: use your graph)?

2. If you plot versus , instead of

versus , how would the graph look like?

(Make a sketch). How would you calculate, from this graph, the acceleration and the speed when the ball crosses the first sensor?

3. In the actual experiment, in the surrounding air, if you use a very light ball (same size), would you get the same value of ? Why?

4. We neglected during the experiment the force exerted on the ball by air resistance. Does this cause a random error or a systematic error? How does it affect the measured value of ?

5. This experiment is quite simple and seems to be accessible even to people many centuries ago. Which measurement in the experiment do you think was difficult to achieve with any acceptable accuracy at that time (i.e. centuries ago)? Can you think of a way to reduce this difficulty? (Galileo did!).

26

Understanding Motion

Experiment (2)

1 Aim of the experiment You develop in this experiment a sound understanding of the different components of motion (position, speed and acceleration).

2 Introduction We say that an object is at rest (i.e. not moving) when its distance to a fix reference point (also called

origin) does not change when time goes on. We can write this statement symbolically as follow:

(1)

where is a constant with the proper units (meter say). This equation says that the object-origin

distance, or simply object’s position, is the same constant at all times . We can also represent

this statement graphically:

Figure 1: graph representing an object at rest at

We have plotted the object position the Y-axis and the time on the X-axis. The graph, as was the case for

Eq. 1, says that the position is the same (i.e. constant and equals ) at all times .

We see, however, in everyday experiment that not all objects are at rest and some of them do move, i.e.

their positions change with time. In this experiment, we restrict our discussion to the linear motion only.

That is to objects which move in a straight line. We shall discuss two cases: in the first case the objects

move with constant speed, and in the second they do so with constant acceleration (i.e. uniformly

increasing or decreasing speed). We start with the first case.

27

The speed21 is defined as the rate of change of position with respect to time. The speed is constant

when this rate is constant i.e. when the ratio of the change of distance to the corresponding

interval of time is constant:

(2)

where is a constant with the proper units (

say). This equation means that the position changes

uniformly with time:

(3)

If we designate the initial distance (i.e. at ) by then and Eq. 3 can be written

as which in turn can be put the following form:

(4)

Notice that Eq.4 is the equation of straight line in the plane with a slope and intercept given by

and , respectively.

Equations 2 and 4 can be represented graphically by figure 2. The figure on the left says that the speed is

constant at all times. The one on the right says that the position of the object increases linearly with

time.

Figure 2: graphs representing an object moving with constant speed

The rate of change in the speed with respect to time is called acceleration. The acceleration is constant

when this rate is constant, i.e. when the ratio of the change of speed to the corresponding

interval of time is constant:

(5)

Where is a constant with the proper units (

say). Equation 5 implies that the speed changes

uniformly with time:

21

Since we are considering only linear motion, there is no need to make any distinction between speed and velocity.

28

(6)

Similarly to Eq. 4, this last equation can be written as:

(7)

Where is the value of the speed at . It can be shown22 that the position of the object is given by

the following equation:

(8)

We can, as we have repeatedly done before, represent Eqs. 5,7 and 8 graphically. The graph on the left

shows the constant acceleration, the one in the middle represents the linearly changing speed and the

graph on the right side shows the quadratically changing object-origin distance (or position).

Figure 3: graphs representing a object moving with constant acceleration

Remember that, following Newton, a constant acceleration is produced by a constant net force. In

general, however, the acceleration may not be constant (when the applied force is changing). But in

many practical applications we can consider with very good approximations the applied forces, and thus

the acceleration, constant so that our previous discussion is quite general.

You will build in this experiment a real feeling of what do the equations and curves developed in this

section mean.

3 Equipment needed : Motion sensor, Computer and Pasco interface.

22

The average speed is defined by

which can be written as . On another side, since,

in the case of constant acceleration, the speed changes linearly, the average speed between two instants (0 and )

will be the speed midway between the initial and final speeds

. This means

. Using

the fact that we get

. (See Giancoli page 28). See the

footnotes in the previous experiment for an alternative derivation (using integration).

29


4.1 Constant position graphs 1. Start Data Studio program. 2. Select from the list of sensors the Motion sensor and double click on it. You should see

then that the sensor is wired to the interface box. 3. Click on position (on the top left of the screen), drag and release it on the graph

icon. A new frame will pop up; enlarge it. 4. Do the same for velocity and acceleration. You should end up with a frame containing

three graphs: position, velocity and acceleration.

5. Draw using the pen button your prediction on the position graph. The prediction graph should look like Fig.1 in the introduction (see above), i.e. representing constant position. Choose a practical distance which you can realize in the lab.

6. Try to reproduce this graph using the motion sensor: when you click on start button

, the computer will start plotting your position in function of time. The plotted graph should nearly superpose you prediction.

7. Make sure the corresponding graphs you get for the velocity and acceleration make sense to you.

8. Save your data for your later use. (Do not forget to delete your files at the end of the lab).

4.2 Constant speed motion 1. Your task in this section is to reproduce a constant speed graph.

2. Draw using the pen button your prediction on the velocity graph. The prediction graph should look like the graph on the left of Fig.2 in the introduction, i.e. representing a constant speed. Choose a practical low speed which you can realize in the lab.

3. Try to reproduce this graph by walking in front of the motion sensor: when you click on

start button , the computer will start plotting your speed in function of time. The plotted graph should nearly superpose your prediction.

4. Make sure the corresponding graphs you get for the acceleration and position make sense to you. (See figure 2 in the text).

5. Save your data for later use. (Do not forget to delete your files at the end of the lab).

4.3 Constant acceleration motion 1. Your task in this section is to reproduce a constant acceleration graph. This may reveal

itself a bit difficult. Try to walk away from the motion sensor while increasing your pace

(speed).

30

2. Draw using the pen button your prediction on the acceleration graph. The prediction graph should look like the graph on the far left of Fig.3 in the introduction, i.e. representing a constant acceleration. Choose a practical low acceleration which you can realize in the lab.

3. Try to reproduce this graph using the motion sensor: when you click on start button

, the computer will start plotting your acceleration in function of time. The plotted graph should nearly superpose your prediction.

4. Make sure the corresponding graphs you get for the speed and position make sense to you. (See figure 3 in the text).

5. Save your data for later use. (Do not forget to delete your files at the end of the lab).

5 Data Analysis 1. Explain the curves you obtained in the three activities. Use the equations developed in

the introduction. 2. For each activity write down, when possible, the values of position, speed, acceleration.

6 Further questions 1. Prove Eq. 8. (see footnote 18 on page 25). 2. Can you prove

, where , and , are the final and initial

speeds and positions, respectively. Hint: eliminate the time between equations 7 and 8. 3. The objective of this experiment is to understand the components of motion (position,

speed and acceleration). Make sure you have reached this objective.

31

Newton's Second Law

Experiment 3

1 Aim of the experiment In this experiment you will investigate the relation between force and acceleration and thus

test Newton’s second law.

2 Introduction Newton’s second law says that the acceleration of an object is proportional to the sum of the

forces applied on that object:

∑

Figure 1: Newton's law experiment setup

We apply this law to our system (chart and hanging mass) shown in figure 1. First we draw the

free body diagram for each object as shown in figure 2.

Figure 2: Free Body diagram for cart and hanging object

32

We then apply Newton’s second to each object neglecting the mass of the string and the pulley

and choosing the downward direction as the positive direction.

(1)

(2)

We omitted for simplicity the vertical projection for the cart. Because the two objects are

connected together through an inelastic string they both have the same acceleration . The

neglecting of the mass of the string implies that the tensions at its ends are equal as it can be

seen by applying Newton’s second law to the string itself23:

Or

(3)

Using Eqs. 1-3 we get

And thus

(4)

Note that if , ( in this case of course) then , i.e. object 2 will have a

free fall motion. Note also that if (here too there is no friction) then i.e. object 1

will stay at rest on the track.

Because the acceleration , Eq. 4, is constant, the distance covered by the cart depends

quadratically on time (see Giancoli, Chap. 4):

(5)

We can write this equation in a slightly different form:

(6)

23

To be precise we should use the forces applied on the string itself which we may call and . But a simple analysis shows that we do have in fact and . (Remember that and are applied on the chart and hanging mass, respectively).

33


intercept This means that if we plot

against (i.e

vs ), we get a straight line24 from

which we can calculate the slope and thus the acceleration (Note that from the above

equations

.

3 Equipment needed Two photo-light sensors connected to a digital timer, pulley, track &cart, masses .


1. The setup should be already connected as shown in Fig.1. If this is not the case talk to the instructor.

2. Check that the digital timer is working fine by passing your hand through the first sensor, to start the time (this is t=0), and then through the second sensor, to stop the timer. The time elapsed between starting and stopping the timer will be displayed on the timer screen.

3. Estimate the amount of static friction exerted on the cart by the track. You can do that in the following way:

a. Remove the hanger and put a so small mass of paper so that the cart does not move. This means that the friction is bigger than the weight corresponding to .

b. Keep adding small masses (of paper) until the cart starts moving. At this moment we can say that the friction is smaller than the weight of object 2.

. This gives us then an upper limit on the friction. c. Weigh the total mass of paper and record it.

4. Now we want to use a mass so that we can neglect the friction besides the weight (i.e. ). A factor of 20 may be enough, so estimate such that

. Put this mass on the hanger.

5. Measure the mass of the chart and estimate its error. 6. Set a distance of 25 cm between the two sensors (the distance should be measured, say,

from the middle of one sensor to the middle of the other sensor). Do not forget to estimate the error on your measurement.

7. Choose a starting point for the cart, near the end of the track. Mark this point with a pencil so that you can always start the cart from this same point. (Why this position must not change?). Hold the cart steady at , and then release it.

8. Record in the table below the time it takes the cart to cover the distance between the two photo-gates. Repeat this measurement two more times.

9. Repeat steps 5-7 for the distances shown in the table.

24

After the fitting procedure.

34

Distance between the

two sensors

Time Average

Time

25

30

35

40

45

50

5 Data Analysis 1. Estimate the errors on and calculate, using the standard deviation method,

those on . (use the row that would give you the biggest error).

2. Calculate the error on

.

3. Plot, using Excel,

versus time (i.e. on the X-axis, and

along the Y-axis) and

draw the error bars (on the same graph). Do not forget to put the titles and units on

the axes.





8. Calculate the (experimental) acceleration and its error (use the fit slope and its error).

9. Calculate the predicted (theoretical) acceleration using Eq. 4 (neglect the friction term).

10. Calculate the error on the predicted acceleration. You should estimate the error on the

different masses and propagate them to the acceleration.

11. How does the experimental acceleration compare to the theoretical one?

6 Further Questions 1. What are the effects of friction and air resistance, i.e. how do they affect your value of

the measured acceleration?

2. You did estimate the upper limit of the static friction; by how much you expect it to

change your value of the measured acceleration? (the kinematic friction is always less

that the static one).

3. How does the acceleration obtained in this experiment compare with ? Does is make

sense to you?

35

Measurement of Viscosity of a Liquid

Experiment (4)

1 Aim of the experiment In this experiment you measure the viscosity coefficient of an engine oil.

2 Introduction The viscosity coefficient quantifies the daily experimental fact that different fluids (i.e. liquids

or gazes) present different friction or drag to moving bodies. A stone, for example, moves faster

in water than it does in oil. More precisely, for small enough spheres moving with small

velocities, the drag force is give by Stokes’ law25:

(1)

Where is the radius of the sphere, the viscosity coefficient of the fluid and the velocity of

ball in the fluid. It’s important to notice the linear dependence of Eq. 1 on the velocity. This

means that this force is zero for a static, i.e. not moving, object and does increase, linearly, as

the object accelerates. The minus sign means that the force is in the opposite direction of

motion, so it’s a resisting force trying to stop the object.

Consider now the setup shown in figure 1: A small sphere released from above a liquid is

moving through it.

Figure 1: Viscosity measurement setup

25

Stokes’ Law is a solution of the Navier-Stokes flow equation for small spherical object moving with small speed. See any Book on Hydrodynamics (e.g.: T. E. Faber, Fluid Dynamics for Physicists).

36

The free body diagram in this figure shows that three forces act on the ball:

1. The gravity force exerted by the earth on the ball. Near the earth surface this is

given by or

where and are the density and

radius of the sphere, respectively, and the gravity acceleration (where we have used

the fact that the mass of the sphere

).

2. The buoyant or Archimedes force exerted by the liquid on the ball. It is due the

fact that the pressure, thus the force, in a liquid increases with the distance from the

top surface of the liquid: the force applied on the lower area of the sphere is bigger than

the force applied on the upper area, resulting in a net force on the sphere (opposite to

the gravity force). The magnitude of this force given by26

where is the mass of the displaced liquid and is equal

( is

the volume of the displaced liquid and is, obviously, equal to the volume of the spherical

ball).

3. The third force is Stokes’s force mentioned above.

The motion of the sphere under the action of these three forces consists of two different parts:

In the first part, the ball undergoes a uniformly accelerated motion. This is because, initially,

Stokes force is quite small and the gravity force outbalances both the buoyant and Stokes’s

forces. The ball being accelerated sees its speed increasing and thus encounters an equally

increasing drag force (recall, Eq. 1, that the drag force is proportional to the speed). It comes a

time then when the sum of the buoyant and Stokes’s forces exactly balances the gravity force.

At that moment the net force is null and therefore the acceleration is also null; the speed of the

ball remains then constant. This in turn causes the drag force to stay constant and consequently

the ball’s t motion enters a constant speed regime. This is the second part a shown in Fig. 1.

This constant speed is called the terminal velocity and can be calculated as follow: At terminal

velocity we have the sum of the buoyant and Stokes’s forces equal to the gravity force:

(2)

Or

(3)

Which is equivalent to:

(4)

26

See Giancoli chapter 17, page 348.

37

And finally

(5)


intercept This means that if we plot against (i.e vs ), we get a straight line27 from

which we can calculate the slope and thus the coefficient of viscosity (Note that from the

equations above

where all quantities are know but ).

3 Equipment needed: Glass tube full with engine oil, Micrometer, Meter stick, Spherical balls, Stop watch, Magnet

and sticky tape.


4.1 Part 1: A quick check for the constancy of the terminal velocity. 1. Put, using sticky tape, two marks on the tube about 30 cm apart. 2. Take the smallest ball (which by Eq. 5 have the smallest terminal velocity), release it

from a fix distance above the tube, record the time it takes to cross the marked distance and fill in the table below. Repeat this measurement 3 times. (Do not change the launching point!). You have to extract the ball from the oil using the provided magnet.

3. Repeat the steps above for distances of 40 and 50 cm.

Table 1: speed vs distance

Distance (cm) Speed Average speed

30

40

50

4.2 Part 2: measure the coefficient of viscosity

1. Keep the two marks at 50 cm apart. 2. Measure the radius of a given ball and measure the time it takes to cover the distance

between the two marks. Do not forget to estimate the error on the distance and time. 3. Repeat the same measurements for the other balls and fill in table 2.

27

After the fitting procedure.

38

Table 2: speed vs radius

Ball Radius Speed Average speed

5 Data Analysis

5.1 Part 1: Constancy of the terminal speed. 1. Calculate, using the standard deviation method, the error on the speed. (Use the row

that would give you the biggest error). 2. Is the terminal velocity constant (as the theory predicts)? (i.e. Is the speed the

same, modulo your errors, in the three trials?)

5.2 Part 2: measurement of the coefficient of viscosity. 1. Calculate the error on the speed (either using error propagation or standard

deviation).

2. Plot, using Excel, versus (i.e. on the X-axis, and along the Y-axis) and draw

the error bars (on the same graph). Do not forget to put the titles and units on the

axes.





7. Calculate the (experimental) viscosity and its error (use the fit slope and its error). You

are given and .

1. The textbook value for the viscosity of Engine oil is (see Giancoli, page 359) . How does your result compare with it. (You have to convert your units for (i.e. ) to (Pascal x second)).

6 Further questions 1. How does the coefficient of viscosity depend on temperature (does it increase or

decrease with increasing temperature)?

39

2. In a pipe and under the same pressure, who runs faster: the liquid with high viscosity coefficient or the one with a lower viscosity coefficient?

40

Conservation of Mechanical Energy

Experiment (5)

1 Aim of the experiment You investigate in this experiment the conservation of mechanical energy.

2 Introduction We deal in this experiment with two forms of energy: Potential and Kinetic energy. Consider

the setup in figure 1:

Figure 1: Transforming potential energy to kinetic energy

The hanging object of mass has a gravitational potential energy given by (see footnote28):

(1)

Where is the object-floor distance and the gravity acceleration. When we let the object fall,

the potential energy of the hanging object decreases, because decreases in Eq. 1. This lost

potential energy is transformed to translational kinetic energy for the object itself and to

rotational kinetic energy for the platter. (Because some of the energy is imparted to the disk, 28

The potential energy possessed by an object at height is given by the work needed to rise this object from the floor (a position at which we consider the object having zero potential energy) to the considered height. This work

is done against the constant gravitation force (see Giancoli, Chap. 8): ∫

∫

.

41

the hanging object has less kinetic energy that when it has a free fall). The translational kinetic

energy of an object of mass moving with speed is given by (see footnote29):

(1)

In a similar way the rotational kinetic energy of an object with moment of inertia rotating with

angular speed is given by (see Further Questions below):

(2)

At the moment just before releasing the hanging object our system (mainly object plus platter)

has only potential energy (all parts are at rest, so there is no kinetic energy):

(3)

After the hanging object is released and has travelled the distance , the object itself and

platter would have acquired a linear and angular speed and , respectively, such that the

conservation of energy must hold true:

(4)

(remember that we took the gravitational potential energy to be zero at the floor level).

Because the object and the platter are connected through a thread, their respective speeds are

related: when the platter rotates with angle , the object moves a distance

where is the radius of the pulley (see Fig. 2 and 3 ). It follows that if changes with time with

a speed

, then the distance changes with a speed

. Thus

object linear speed and the platter angular speed are related by .

We can then eliminate from Eq. 4 and get the following form:

(5)

29

Imagine that some net force is used to accelerate the object from rest ( to a speed . The work done by

this force is ∫ ∫ ∫

∫

∫

. (Where we have used Newton’s

law:

).

42

Figure 2: relation between linear and angular displacements

We can easily see, from the last equation, that knowing the hanging object mass and the

platter inertia , all we need to test Eq. 5 is to measure and . We can then plot versus

and should obtain a straight line with a slope equal one and zero intercept. This is

one the of tasks in this experiment.

As it was just said we need to know the moment of inertia of the platter . We can obtain in

two ways:

1. Since the platter can be approximated by a flat disk, we can calculate using the

formula which gives the moment of inertia of a disk of mass and radius :

(6)

Hence by measuring and we can find the moment of inertia of the platter.

2. The second method to measure is by using Newton dynamics. Consider the free body

diagram, figure 3, corresponding to our system.

Figure 3: Free body diagram of the platter and hanging object system

The application of Newton’s laws for translation and rotation to the object and platter,

respectively, gives the following two equations30:

30 We assume an inelastic massless thread, which means that the same tension exists on both ends (applying

Newton’s second law to the thread we get i.e. ). We neglect also the mass

of the pulley and thus no or very small energy is wasted there.

43

(7)

(8)

where and are the torques due to the tension and frictions, respectively. Here

as was the case for speeds, the linear and angular acceleration are related: .

Please note that here refers to the radius of the pulley and not that of the platter (see

Fig. 3). We can extract from the first equation and substitute it in the second (we also

eliminate using ):

(9)

(10)

The last equation can be rearranged in the following form:

(11)

It can be easily seen that represents the slope of the line obtained by plotting

in function of . Therefore we can measure by changing , measuring

corresponding linear acceleration and then graphing Eq. 11.

Both methods for measuring the moment of inertia of the platter will be used in this

experiment.

3 Equipment needed : Platter, smart pulley connected to computer, set of masses, balance, Vernier caliper, meter.

4 Experimental procedure 1. Measure the radius of the platter and estimate their errors. Do not remove the platter. 2. The mass of the platter is give to you . Record it on your logbook. 3. Measure the radius of the pulley over which the thread is winded. Estimate its error. 4. Measure carefully the distance which the object could travel and estimate its error (see

Fig. 1). This is basically the distance from the lower end of the pulley to the floor (If not sure talk to the instructor).

5. Before starting taking the remaining needed data, make sure the platter is set perfectly horizontal, the pulley and the thread perpendicular the radius of the platter and that every part can move without too much friction.

6. The pulley is connected to a computer (using Pasco interface) which can display the object instantaneous speed and acceleration. Here is what you need to do:

44

a. Start Data Studio program.

b. Select from the list of sensors the Smart pulley and double click on it. You should

see then that the sensor is wired to the interface box.

c. Click on velocity (on the top left of the screen), drag and release it on the graph

icon. A new frame will pop up; enlarge it.

d. Press the start button and make sure the computer records tangible data

when the pulley rotate (you can rotate the pulley with your hand to test the good

working of the computer-sensor system).

e. You can read the final maximum speed and by highlighting the corresponding

curve and pressing the button.

f. To get the acceleration, use the slope button . Put the slope line on the linear

part of the speed curve in order to get a more accurate measurement of the

acceleration.

7. Put on the hanger (do not forget to consider its mass) in turn each of the masses given in

the table below and measure the maximum speed and acceleration of the object after having travelled the distance (while the acceleration of the object should be constant, its speed, of course, increases with time.) Fill in the table.

(

) (

) (

) (

)

0.080

0.110

0.140

0.170

0.200

5 Data Analysis

5.1 Part 1: measuring the platter moment of inertia 1. Calculate, using the standard deviation method, the errors on and .

2. Plot versus using Excel and draw the error bars. Do not forget the units

and titles. Remember that here is the radius of the pulley and not the radius of the

platter.




6. Plot your data, error bars, fit line on a graph paper this time (do this for part 1 only).

45

7. Calculate the experimental values of , . This is the first method for measuring .

8. Calculate, and this is the second method, the theoretical values of , using Eq. 6

where is the radius of the platter.

9. How do the two values of compare?

10. What can you tell on the frictions in this experiment (see Eq. 11)?

5.2 Part 2: Conservation of energy

1. Plot, using Excel, versus

and draw the error bars. Do not forget the

units and titles. ( is the radius of the pulley). Please note that the variable on the x-axis contains , the moment of inertia of the platter, and not the number 1.

2. Fit linearly your data and get the slope and the intercept and their errors. 3. Does your data represent the theory given by Eq.5? Look at the fit quality variable . 4. Do the slope and intercept have the expected values? What do you conclude about the

conservation of energy. 5. What does a significant non-zero value of the intercept mean?

6 Further questions 1. Show that the rotational kinetic energy of an object with moment inertia rotating with

an angular speed is given by

. Hint: use the definition of the work done

by a torque ∫ and follow the same steps as in footnote 2. 2. Suggest how the speed and acceleration are measured in this experiment?

46

Hooke’s Law and Harmonic Motion

Experiment (6)

1 Aim of the experiment In this experiment you test Hooke’s law and explore harmonic oscillations.

2 Introduction The spring (Fig. 1-a) is an example of an elastic body which exerts a restoring force when

stretched or compressed. This force is given by Hooke’s law31:

(1)

Where gives the magnitude and direction of the spring deformation and is a proportionality

constant measuring the stiffness of the spring (a high value of means that the spring is harder

to stretch). The minus sign in Eq. 1 means that the force is directed in the opposite direction of

the displacement and tends to return the object to equilibrium position.

Figure1: spring elongations

When an object is hanged to a spring, with a given , the latter stretches a distance

just enough for the restoring force to counter the weight of the object (Fig.1-b) and reach a

static equilibrium. Written in symbols this means:

(2) 31

This is valid only for small deformations.

47

where is the magnitude of force applied on the object by the earth and is the

magnitude of force exerted on the object by the spring (the two forces have, of course,

opposite directions).

If we pull the object down a distance from its equilibrium position (Fig.1-c) and then release

it, the object sets up to oscillate. To study this motion we apply, as usual, Newton’s second law

to the object:

where

is the object acceleration. We consider the instant just after releasing the object

and we choose the positive direction downward as shown in Fig. 1. Projecting the previous

equation in the chosen vertical axis we get: 32

(3)

This equation can be re-written as:

The first two terms represent the equilibrium condition (Eq. 1) and therefore cancel out. We

are then left with the following equation:

Which is equivalent to:

(4)

This is an ordinary linear differential equation33 where the unknown is the function of time

. It’s easy to guess for a solution to this equation by noticing that any function which has its

second derivative proportional to the function itself verifies Eq. 4. The sine and cosine

functions are examples of such function. It can be shown that the most general solution of Eq.4

which describes the motion of the hanging mass is given by:

32

The acceleration is

and therefore has the same sign (after projection) as . In our case it’s positive. 33

It is an equation which relates the still unknown function and its second derivative.

48

where √

is the proper angular frequency34, and and are constants determined by the

initial conditions of the object (i.e. its position and velocity at ). A such sinusoidal motion

is called a harmonic. The object thus will oscillate with a frequency:

√

(5)

and a period:

√

(6)

We can cast Eq. 6 in a linear form by squaring both sides. We then get:

(7)

This equation takes a linear form if we consider and as the dependent and independent

variables, respectively. It then represents a straight line with slope

and zero intercept.

3 Equipment needed

Coil spring, supports, clamps, slotted weights and weight hanger, Meter stick, Stop watch.


4.1 Part 1: Measurement of the spring constant using statics 1. Mark the position of the end of the spring when it is empty. You will measure the

stretching of the string from this position. 2. Suspend the weight hanger on the spring, record its mass, and read the spring extension

. Write down your values in table 1. 3. Add 0.5 kg to the hanger, record the new total weight and read the corresponding

spring displacement. Repeat this step and fill table 1. Table 1

Object total mass Object weight Spring extension (cm)

1. (hanger’s mass)

1.5

2.

2.5

34

It’s called the proper frequency because it depends on the properties of the spring-object system alone (i.e. spring constant and object’s mass). The system likes to oscillate at this frequency and it will do so whenever it is let free to oscillate.

49

3

4.2 Part II: Measurement of the spring constant using dynamics 1. hang each of the masses listed in table 2 to the spring, pull it down a small distance and

release it (let it oscillate). It’s may be easier for you to start with biggest mass (i.e. )

and finish the smallest one.

2. Using a stopwatch, measure the time of, say, 10 oscillations35 and then calculate the

period (i.e. the time of one oscillation). Fill in table 2.

Table 2

Object mass Time of 10 oscillations

Period

1. (hanger’s mass)

1.5

2.

2.5

3

5 Data Analysis

5.1 Part I: 1. Estimate the errors on mg.

2. Plot, using Excel, mg versus time (i.e. on the X-axis, and mg along the Y-axis) and

draw the error bars (on the same graph). Do not forget to put the titles and units on the

axes.





7. Calculate the spring constant and its error (using the slope from Eq. 2).

5.2 Part II: 1. Estimate the errors on . (Apply the standard deviation method for a row that would

give you the biggest error). 2. Plot, using Excel, versus (i.e. on the X-axis, and along the Y-axis) draw the

error bars (on the same graph). Do not forget to put the titles and units on the axes.

35

We use such a number of oscillations to minimize the error on the period.

50



5. Does your data represent the model given by Eq.7? Look at the fit quality variable .

6. Calculate the spring constant and its error (using the slope in Eq.7).

7. How this value of compares with the one obtained in the previous section?

6 Further questions 1. For a given mass, if we replace the actual spring with another one with a greater value

of spring constant , do the period get smaller or bigger? Explain.

2. Considering the oscillations of an object hanged to a spring, at what point on its

trajectory does the object has the maximum speed? At what point it has the maximum

acceleration? What is the value of the total force at that point? Is it at its maximum too?

(i.e. is Newton’s second law respected?)

3. If we put the spring-object system, i.e. same spring and object, on an inclined planed

(Fig.2), will it oscillate with the same period as when it is set vertical? Explain. (Assume

that there is no frictions).

Figure 2

51

Simple Pendulum

Experiment (7)

1 Aim of the experiment You study in this experiment the harmonic motion of simple pendulum.

2 Introduction A simple pendulum consists of a mass (bob) and an inelastic massless string (see Fig.1). when

the bob is given a small tilt angle and then left to itself, it starts oscillating. The reason is that

by moving the bob in the earth gravitation field, and thus doing work, we have given it a

potential energy which manifests itself in the kinetic energy of the bob. If the pendulum is

isolated the mechanical energy is conserved and change its form continuously from potential to

kinetic energy and vice-versa.

Figure1: simple pendulum

Let us apply Newton’s second law to the bob and see what kind of motion it undergoes. There

are two forces exerted on the bob: the gravity force exerted by the earth and the tension force

exerted by the string.

(1)

52

We can decompose the gravity force to two components one along the direction of motion and

the other perpendicular to it with magnitude (see Fig. 1) and , respectively.

Because there is no motion, and thus no acceleration, along the string (because its length is

fixed, we assumed that the string is inelastic), Eq. 1 gives us:

(2)

We consider the instant just after releasing the bob and choose the positive direction to be

counterclockwise as shown in Fig. 1. It follows then the second equation of motion,

corresponding to the direction perpendicular to the string, is given by (There is no component

of tension in this direction!)36:

(3)

We can change the linear acceleration to the angular acceleration

using the formula

(because, being the bob’s displacement along the arc, we have , which implies

that

) and get the following non-linear differential equation:

(4)

Starting from this equation one can find, in an non-easy way though (see Appendix C), that the

period is given by the following series ( being the maximum angular displacement):

√

(

(

)

(

) ) (5)

However, for small oscillations, when the angle remains small at all times, one can use the

small angle approximation:

(6)

and simplifies Eq. 4:

(7)

36

The acceleration is given by

and thus has the sign of which is positive in our case (we chose the

counterclockwise as the positive direction).

53

This is a linear easy to solve equation. We are looking, basically, for a function which when

differentiated twice (i.e.

gives the same function . It’s well known that a cosine

function has the required property. The solution to Eq. 7 is then:

(8)

Where is the maximum angular displacement and the angular frequency. The motion of

the simple pendulum is therefore harmonic. Substituting Eq. 8 in Eq. 7 we get:

(9)

The period

is then given by:

√

(10)

This is also what one gets from Eq. 5 for small angles where one neglects the terms containing

(

).

We can put Eq. 10 in a linear form by squaring both sides. We then get:

(11)

This equation is indeed linear if we consider and as the dependent and independent

variables, respectively. It then represents a straight line with slope

and zero intercept.

3 Equipment needed : Small mass, string, Meter stick, Stop watch

4 Experimental procedure 1. Set the length of the pendulum to 30 cm. 2. Give the mass a small (less than ) tilt angle , release it and measure the time of 5

oscillations. (the pendulum should swing in a plane and should not do rotations). Repeat this measurement 3 times and fill in the table below.

3. Repeat the steps above for the other lengths listed in the table.

Table 1: simple pendulum

Length of

pendulum

time of 5 oscillations Average time of 5

oscillations

Period

54

30

40

50

60

70

80

5 Data Analysis 1. Estimate the errors on . (Apply the standard deviation method for a row that would

give you the biggest error). 2. Plot, using Excel, versus (i.e. on the X-axis, and along the Y-axis) and draw the

error bars (on the same graph). Do not forget to put the titles and units on the axes. 3. Fit linearly your data and get the slope and the intercept.


5. Does your data represent the theory given by Eq. 11? Look at the fit quality variable .

6. Calculate the gravity acceleration and its error (using the slope in Eq. 11).

7. How this value of compares with the text book value of ?

6 Further questions

1. What should you plot on the graph axes so that the slope is equal to g ?

2. If one uses a tilt angle , what is the difference between the periods given by Eq. 5 and Eq. 10. (use only the first two terms dependent on in Eq. 5, i.e. neglect the higher terms).

3. Does the period of the pendulum depend on the mass of the bob? Explain. 4. The period of the pendulum is approximately independent of the amplitude. Can you

think of an application where one uses the pendulum.

55

Static Equilibrium

Experiment (8)

1 Aim of the experiment You investigate in this experiment the conditions of static equilibrium.

2 Introduction An object is in static equilibrium when it position and orientation, with respect to a reference

frame, does not change. Following Newton, this happens when the sum of applied forces and

torques equals zero. We can introduce the notion of torque in the following way. Consider

figure 1.

Figure 1: static equilibrium

The hanged object is static equilibrium because the applied net force is zero: the gravity

attraction force is balanced by the thread tension. In this case both forces act along the same

line and the equilibrium condition simply reads:

(1)

where and are the magnitudes of the weight and tension forces, respectively. Now

consider figure 2 where we have two objects of different weight on a see-saw. The two objects

sit at different distances from the center (pivotal point). Day life experiments show that the

equilibrium condition is not where and are the objects respective weight but it is

given by:

56

(2)

where and are the position of the objects from the center of the see-saw. and are

also called lever arms. Therefore, in this case, it is the product (force times lever arm) which

is of importance. It’s called the torque of force .

Figure 2: static equilibrium condition in a see-saw.

It’s very important to notice that represents the perpendicular distance, i.e. the smallest

distance, to the force and that the torque is calculated with respect to a fix pivotal point (here

the center of the see-saw). For example the torque37 of the force shown in figure 3 is

because the perpendicular distance from the force to the pivotal point O is

and not .

Figure 3: Torque definition.

This is equivalent to decomposing the force into its parallel and perpendicular components

(see figure 3) and considering the torque of each. The parallel component, to the bar, has zero

torque (because its lever arm is zero since it passes through the pivotal point) . The

perpendicular component of magnitude has a torque equal to (since its

37

See Giancoli page 256 for more explanation.

57

lever arm is ) 38. The net total torque is then the same as found previously

.

In summary, the two conditions of equilibrium are as follow:

∑

∑

We will apply these conditions to the system in our experiment. Figure 4, representing the

setup of the experiment, shows a horizontal bar of length which is free to rotate around a fix

point O. The other extremity of the bar is connected to a force sensor and a thread which keeps

the bar in a horizontal situation. A constant weight is hanged on the bar at a distance from its

pivotal point O.

Figure 4: Experiment setup.

Since the bar is in equilibrium, it means that he sum of torques with respect to O is null. We do

actually have two torques acting in opposite directions:

The first one, , is due to the weight of the hanged mass with a lever arm equal

to . The due torque has then a magnitude .

The second torque, , is due to tension of the thread with a lever arm equal to .

The tension is, of course, along the thread which makes an angle with the bar. The

resultant torque is then (see discussion relative to figure 3).

The equilibrium condition reads:

38 Other considerations show that the torque can be defined as a vector quantity whose direction and magnitude

is given by the vector cross product: .

58

(3)

or

(4)

The last equation can be re-written as

(5)

which is a linear relation between and . The slope is therefore

and the intercept is

zero.

3 Equipment needed : Force sensor, Computer, Pasco interface, Stand, thread and bar, ruler, masses and mass holder.

4 Experimental procedure 1. Make sure that the stand, bar and thread make a right angle triangle as show in figure

4. 2. Measure the length of the horizontal bar and estimate its error. 3. Measure the angle and make sure it stays fix during the whole experiment. Estimate

the error on your measurement. 4. Start the Data Studio program and double click on the force sensor (the one with a

hook) from the list of available sensors. You should see then that the sensor is wired to the interface box.

5. Click on graph icon and enlarge the pop up window. 6. Calibrate the force sensor as follow:

a. With no weight on the horizontal bar, push start on the Data Studio program to measure the tension applied on the sensor. Press stop after a while.

b. Press the tare button localized on the force sensor itself. c. Press start again and after a while press stop. The recorded tension will be your

zero. In other words, you should add the actual reading to all other readings. 7. Put a constant mass at a distance from the pivotal point O. Do not forget

to record the value of the mass .

8. Press start, wait for a short moment and then press stop. Use the button and read the mean value of the tension. Record your reading in the table given below. Do not forget to estimate the error on the tension (you may want to repeat a measurement three times and use the standard deviation method to calculate the error).

9. Repeat steps 7-8 for the other distances listed in the table.

59

50

40

30

20

10

5 Data Analysis 1. Write down here the error on .

2. Plot, using Excel, versus time (i.e. on the X-axis, and along the Y-axis) and draw

the error bars (on the same graph). Do not forget to put the titles and units on the

axes.





7. Calculate and its error (from the slope and its error).

8. Calculate and its error from the angle and its error you did measure at the

beginning of the experiment.

9. How do your values for compare to each other. Do they agree? What do you

conclude about the equilibrium condition (Eq. 4).

6 Further questions 1. It was said in a footnote in the introduction that the torque due to a force with a lever

arm can be written as a cross product . The magnitude of is

where is the angle between and . How is the direction of the torque defined?

2. In the introduction we evaluated the involved torques with respect to the point O and arrived to Eq. 5. We can arrive at the same result (i.e. Eq. 5) by evaluating the torques with respect to the point E. Here are the required steps:

a. List the torques in action when we consider the pivotal point at the right extremity E of the horizontal bar (see figure 4). Hint: draw all the forces applied on the bar and consider only those with a non-null torque. Remember that a force passing through the axis or point of rotation has a torque equal to zero.

b. Write the equilibrium condition applicable to the previous case. How does it compare with Eq. 3 (or Eq. 4)?

60

61

Sound Waves

Experiment (9)

1 Aim of the experiment You investigate in this experiment sound waves and the resonance phenomenon.

2 Introduction Although it is sound waves that will be studied in this experiment, we will have a look at waves

on a string first, because it is easier to understand some of the waves properties there. Imagine

a very long string (Fig. 1). When one starts pulling up and down on its near end, a travelling

wave, made of pulses, sets up. Each point of the string oscillates (up and down) with constant

frequency and amplitude (the amplitude and frequency of the moving hand say).

Figure 1: travelling waves on a string

Imagine now that the string is not infinitely long and that its right side is tightened to

something. Then when a pulse arrives at that end it gets reflected. Here we distinguish two

cases: fix and loose end.

2.1 The right end of the string is fix When the pulse arrives at the fixed end it get inverted and reflected back (if, say, it is a

pulse up it becomes a pulse down). This is shown in figure 2. This happens because of

the action-reaction between that end of the string and the object (the wall for example)

that holds it fix. The pulse acts on the wall and the wall act back on the string, creating

an inverted pulse.

62

Figure 2: pulse inversion and reflection at a fixed string end

If we hold tight the string on the left end too, the pulse will be inverted and reflected at

this end too and the motion continue eternally (This is because the energy is conserved

if there is no frictions of course). Please notice that this second reflection puts the pulse

right up as it was created first. When we have many pulses, the ones propagating to the

left can then interfere with those propagating to the right. The interference however

can be constructive only for very specific frequencies . these are called the resonant

frequencies. We can find them by the following argument: Consider the pulse in figure

2, where we assume that the string has a length . To travel from the source of vibration

on the left to the fixed point on the right and back to the vibration source, it takes the

pulse a time

, where is the speed of the pulse on the string (which depends on

the string mass and tension only). To obtain constructive interferences the time

should be an integer multiple of the period of the vibrations source (This way a new

pulse from the source will just add up to the reflected one when it is inverted and

reflected by the source. This is also why the oscillations energy builds up in the string

and thus vibrate with greater and greater amplitudes even though the source amplitude

itself is small). The resonance condition is then:

(1)

Where is a positive integer. Because

, this last equation can be re-written in the

following form:

(2)

This means that at resonance the length of the string is an integer multiple of half the

wave length. At such frequencies the points on the string oscillate with same common

frequency but with different maximum amplitudes. Figure 3 shows the first three

resonances. Please note that it’s the maximum amplitudes which are drawn in this

figure (and other similar figures) and not a snapshot of the motion of the string. For

example when there are two locations of the string, besides its two extremities,

which do not move at all. There, the maximum amplitude is zero and. They are called

nodes. On the other side there are three locations which oscillate with the biggest

63

amplitude possible. They are called antinodes. The remaining points of the string vibrate

with amplitudes between these two extremes39.

Figure 3: first three resonances of a string with fixed ends

2.2 The right end of the string is loose If the right end is loose (not fixed), the pulse get reflected but not inverted (Fig. 4).

Figure 4: reflection, without inversion, of a pulse at a loose end of the string

The pulse, here also, has to travel a distance and thus needs a time

to get

back to the source. But because the pulse has not been inverted at the loose end (but it

will be inverted on the source end), to obtain constructive interferences needs to be

equal

plus any integer multiple of T (The reflected pulse needs to be added to a going

down pulse):

(

) (3)

where is a positive integer. This equation is equivalent to (using

):

(4)

Figure 4 shows the first three resonances.

39

It can be shown, see Giancoli page 414, that dependence of the maximum amplitude of vibration of a point

on the string on its location is given by sine function

. It’s this function which is plotted in Fig.

3.

64

Figure 5: first three resonances for a string with one loose end

Now we go back to sound waves. They are conceived either as small oscillations of the air

molecules (there are no sound waves in vacuum), or alternatively as small variation of the

pressure around its atmospheric value. This second view is illustrated in figure 6 where we see

that the pressure changes periodically as we move on through a tube.

Figure 6: A snapshot picture of Sound waves (viewed as pressure waves)

Note that the molecules oscillate in the direction of the propagation of the waves along the tube ( and

not perpendicular (transversal) to the propagation direction as was the case for a string). For this reason

sound waves are called longitudinal waves (waves on a string are transversal waves).

The resonance phenomena seen with the strings occurs also with sound waves in a tube. Here too we

distinguish two cases:

2.3 Open Tube The open tube case is similar to the one of a string with a loose end (the similarity is better seen

by picturing the sound waves as air molecules vibrations. The molecules at the open end are

almost free and thus have the greatest oscillation amplitude). The resonance condition is then

given by Eq. 2:

, (5)

Figure 7 shows the first three resonances. The (maximum) amplitudes shown are those

of the pressure of the air inside the tube relative to the atmospheric pressure (i.e. our

zero pressure is in fact the atmospheric pressure (see Fig. 6)).

65

Figure 7: first three resonances for an open tube. the maximum pressure amplitudes across the tube are shown.

Note that because the tube ends are in direct contact with the ambient air, the pressures at

those location are equal to the atmospheric pressure (which we took equal to zero).

We can re-write Eq. 5 in a way to display a linear relation between the frequency and the

integer (you will see that for an open tube one has to change the frequencies and find out the

different resonances (i.e. values)).

, (6)

If we consider and as the dependent and independent variables, respectively, then

Eq. 6 is the equation of a straight line with slope

and zero intercept.

2.4 Closed Tube This case can be likened to the one of a string with fixed extremities (see above). The

molecules at the closed end cannot move, because of the wall, and therefore the

oscillation amplitude is zero. In the pressure picture this point represents a location

where the pressure marks the greatest change and thus the (pressure) amplitude is

maximum. This is illustrated in figure 8.

Figure 8: first three resonances of a closed tube. Maximum pressure amplitudes across the tube are shown

The resonance condition is given by (see Eq. 4):

(7)

66

Note that takes only odd values. Equation 7 can be written as:

, (8)

Which is, if we take and as the dependent and independent variables, the equation

of straight line with a slope equal

and zero intercept.

3 Equipment needed : Function generator, tube, speaker, microphone, oscilloscope.


4.1 Part I: Closed tube 1. Familiarize yourself with the setup (see Fig. 9). Notice in particular that the function

generator is connected to the speaker: The former generates sinusoidal electric signals and the latter transforms them, through its vibrating membrane, to audible sounds (tones). Notice also that the microphone is connected to the oscilloscope. The microphone transform sound waves (molecules oscillation) to electric oscillating signals which the oscilloscope displays them on its screen.

Figure 9: Sound experiment setup

2. At resonance the molecules have their greatest oscillation amplitude and thus electric signals are also the greatest. Therefore we can identify a resonance by monitoring the amplitude of the electric signal on the oscilloscope: A resonance occurs when the amplitude on the screen is maximum.

3. Note that that you should leave free a small space between the end of tube and the speaker (this end should always be in good contact with the atmosphere, even for the closed tube!).

67

4. Set the generator frequency at . 5. Starting from a position where the piston is the closest to the speaker, move it slowly

away and watch the amplitude of the signal on the oscilloscope. The signal should increase, reach a maximum at resonance and then decrease. Record the position of the piston at resonance (i.e. when the signal is maximum). The first resonance corresponds to , the second one to , and so on (using positive odd integers only). Fill the table below.

6. Continue moving the piston away from the speaker and find the other resonances. Do not forget to estimate the error on

Table 1: Resonances for a closed tube

1

3

5

7

9

Part I: Open tube 1. In this part the length of tube is constant and you should vary the frequency. 2. Measure the length of the tube and estimate its error. 3. Starting from the lowest possible frequency value, increase it slowly until you hit the

first resonance (You should of course watch the signal on the oscilloscope). Record the value of the resonant frequency in the table below. The first resonance corresponds to , the second one to , and so on.

4. Continue to increase the frequency slowly and find the other resonant frequencies. Do not forget to estimate the error on

Table 2: Resonances for a an open tube

1

2

3

4

5

68

5 Data Analysis

5.1 Part I: Closed tube 1. Graph, using Excel, the data in table 1: Plot versus and draw the error bars. Do not

forget to put the titles and units on the axes.




5. Plot your data, error bars, fit line on a graph paper this time (Do this for the first part

only).

6. Calculate the speed of sound and its error (consider also the error from ).

5.2 Part I: Open tube 1. Graph, using Excel, the data in table 2: Plot versus and draw the error bars. Do not

forget to put the titles and units on the axes.




5. Calculate the speed of sound and its error (consider the error from ).

6. Are the two values of the speed of sound and consistent with each other?

7. Calculate the mean value of and and its error (i.e. the error on the mean value).

This represents your best measurement of the speed of sound.

8. how does you measurement of the speed of sound compare with the world know

values of ?

6 Further questions 1. Figure 8 shows how the maximum amplitude of the pressure oscillation changes all

along a closed tube. Sketch how the corresponding molecules oscillation amplitude

varies throughout the tube. (Remember that sound waves can be thought of either as

pressure waves or molecules displacement waves).

2. What are the frequencies which the human can hear? (you can try check out that with

your setup).

3. Can you suggest why a baby can hear high frequencies that an old person cannot?

69

Specific Heat Capacity of a Solid

Experiment (10)

1 Aim of the experiment You measure in this experiment the specific heat of a metal.

2 Introduction When we give the same amount of heat to different solids, liquids or gases their temperatures

change with different amount (for example the change of temperature of wood will be less

than that of iron both having the same mass). This is so because their heat capacities are

different (iron has a smaller heat capacity than wood40).

We can define the heat capacity of an object as the ratio between the quantity of heat

transferred to the object41 and the resulting increase in temperature :

(1)

The units of are . Clearly, as defined above, the heat capacity depends on the

object’s mass: The temperature of 100 kg of a substance will change a little compared to the

temperature of just 1 kg when heated with the same . We need therefore to specify

explicitly the mass for which the heat capacity is measured. This leads us to define the specific

heat capacity as the heat capacity, as defined by Eq. 1, of one unit mass of the considered

object:

(2)

The specific heat capacity of an any object being known, we can calculate the change of

temperature of that object caused by any amount of heat . This is done easily by re-

writing Eq. 2 in the following form:

(3)

40 and . 41

We assume that the phase of the object does not change. For example if the object is a solid it stays solid (and does not change to other phases, e.g. liquid or gas).

70

This equation means that for an object of mass and specific heat capacity , a quantity of

heat will change the temperature of that object by an amount .

On another side the calorie, the unit of heat, has been defined as the heat needed to raise the

temperature of one gram of water by one degree. It follows from this definition and Eq. 3 that

the specific heat of water is

. Knowing , we can proceed to measure the

specific heat capacity for other substances.

Assume we would like to measure the heat capacity of a metal (e.g. copper). We can do so by

the following steps:

1. Put a mass of water in a calorimeter42 made of the considered metal and then

measure the initial temperature of ensemble (i.e. water plus calorimeter) . One of

course should wait for thermodynamic equilibrium before taking the temperature.

2. Heat up a mass of the metal to a high temperature .

3. Mix the heated metal with the water in the calorimeter (see Fig. 1). What happens is

that the metal will lose some of its heat to the water and calorimeter until the three

bodies get the same final temperature . The temperature of the water and

calorimeter will raise ( and that of the heated metal will decrease ( .

Figure 1: merging of a hot piece of metal with water at ambient temperature

4. According to Eq. 3 the quantity of heat lost by the heated metal is

. The quantity gained by water and calorimeter is +

, where

as explained above.

5. Because energy is conserved the two quantities of heat considered in the previous item

should be equal:

( )

From which we find :

42

A calorimeter is a well insulated recipient. It doesn’t exchange heat with its environment.

71

( ) (4)

Once we have measured, in this way, the specific heat capacity of a metal, we can use it to

make the calorimeter. We can then measure of any other substance (solid or liquid especially)

by using the following equation (deduced from step 5 above):

[ ]

( ) (5)

You will use this last equation to measure the heat capacity of some metal.

3 Equipment needed : Hot plate, boiler, balance, calorimeter, water, thermometer, metal.

4 Experimental procedure 1. The experiment setup is sketched in figure 2 below. Make sure that all needed

equipments are available.

Figure 2: Heat capacity measurement setup

72

2. Measure the mass of the calorimeter (i.e. the water recipient) and the water and estimate their respective errors.

3. Measure the mass of the metal and estimate its error. 4. Measure the water-calorimeter initial temperature and estimate its error. 5. Put the metal in the boiler and heat it up to a temperature around . The

thermometer should be surrounded by the metal pieces so that there is a good thermal

contact between them.

6. Extract the metal and merge it with the water, stir up a bit, and measure the water new

temperature .

7. Repeat steps 2-6 with new different masses of water and metal and fill in the table

below.

Table 1:Experiment data

trials

1

2

3

5 Data Analysis 1. For each row of table 1, calculate the specific heat capacity of the considered metal

(using Eq. 5). You are given and .

2. Calculate the error on c using error propagation (on Eq. 5). Do this for only one row. You can ignore, but you have to justify why, the errors on masses.

3. Calculate the mean value of c using the values obtained in the three trials. 4. Calculate the error on c, now by using the standard deviation method on the three

trials. 5. Are the two errors, calculated in two different way, consistent? 6. How does your best measured value of c (the mean value) compare with the textbook

value (you should know the type of used metal and look for the textbook value).

6 Further questions 1. Which of two objects at the same temperature can cause higher burns when you touch

it: the one with the greatest specific heat capacity or the one with the smallest specific heat capacity?

2. Why we do wrap up potatoes with aluminum foils when we want to cook them in an oven?

73

Mechanical Equivalent of Heat

Experiment (11)

1 Aim of the experiment You measure in this experiment the equivalent in calories of one Joule.

2 Introduction Heat is one of several forms of energy. It was first43 thought that Heat is a kind of liquid, called

calorific liquid, which flows from a hot object to a cold one when they are brought in contact.

When the equilibrium is reached, i.e. the two objects have the same temperature, the liquid

stops flowing. Although this view was successful in explaining many of Thermodynamic

processes it could not give a satisfactory explanation of why frictions do create heat. The

analysis of such phenomena and especially the study of gazes (kinetic theory of gases) revealed

that heat is, in fact, just the random motion of the constituents of mater (atoms and molecules)

and that temperature is related to their translational motion: The average speed of the moving

constituents of a hot object is bigger than that of a cold object. Therefore when one heats up a

body one is, in fact, increasing the speed of the atoms of that body.

Now, if Heat is just the motion of atoms and molecules, which possess then mechanical kinetic

energy, then the unit of heat44 (calorie) should be the same as the unit of mechanical energy

(joule)45. Therefore one can write:

(1)

Where is a dimensionless constant of proportionality to be determined by the experiment.

43

We are roughly speaking about times before 1850’s. 44

See the previous experiment “Specific Heat Capacity of Solids”. 45

Just like the mile and km are the units of length.

74

Figure 6: Joule's experiment setup

James Prescott Joule46 did, in 1843, an experiment where he measured the mechanical equivalent of

heat constant . In his setup, see Fig. 1, a falling known mass generates heat, by friction, in a known

mass of water. The heat generated is, by conservation of energy, equal the potential energy lost by the

falling mass. The heat being measured in calories and the potential energy measured in joules one could

easily measure .

Figure 2: Joule's Experiment setup (electrical version)

In this experiment you proceed in a similar way but instead of using the mechanical way (i.e. a

falling mass) you use an electric power supply to produce heat (see Fig. 2). We know, on one

side, from elementary electricity that the energy released during a period of time by a

resistor when a current flows throw it and a potential is maintained across its ends is given

by :

(2)

If , and are measured in SI units (i.e. ) then the energy is given in

. The released energy or heat, on the other side, causes a rise in

temperature of the surrounding water and calorimeter47 ( and are the initial and final

46

See http://en.wikipedia.org/wiki/James_Prescott_Joule 47

A calorimeter is a well insulated recipient which serves to measure energy (Fig.2 does not show all its features)

75

temperatures, respectively). The measurement of this change in temperature can be used to

measure the released heat by using the following formula (see the previous experiment):

(3)

Where and are the masses of water and calorimeter, respectively, and

and their respective specific heat capacities. If the factors on the right side

hand of Eq. 2 are given in SI units then Q will have calories as units.

Conservation of energy implies that the quantities given by equations 1 and 2 and expressed in

the same units are equal. We can use joules as the one unit by converting calories to joules

using Eq. 1. We get then:

[ ]

(4)

It can be easily seen from this equation that if one plots versus

one then gets a straight line whose slope is equal to the

mechanical equivalent of heat constant , resulting therefore in the measurement of the

wanted value of .

3 Equipment needed : Power supply, calorimeter, water, resistor, ammeter, voltmeter, thermometer, stopwatch.

76

4 Experimental procedure 1. Measure the mass of calorimeter (i.e. water recipient) when it is empty. 2. Put some water in the calorimeter (the water should fully cover the resistor) and

measure the new mass. Deduce the mass of the water. 3. Connect the circuit shown in figure 1 (if not sure, talk to the instructors). Insert the

thermometer in the calorimeter so that you can read the temperature of the calorimeter-water system.

4. Switch on the power supply and set a current of . 5. Record the corresponding voltage and estimate the error on both voltage and current. 6. Record the initial temperature and start the stopwatch as the same time. should

then be the temperature at . 7. Then every 5 minutes record the temperature, the voltage, the current and fill in the

table given below. (You should stir the water from time to time to make sure the temperature inside the water is homogenous). Do not forget to convert all units to SI units (e.g. minutes to seconds).

Time Temperature

0

5

10

15

20

5 Data Analysis 1. Record down here your estimates of the error on , and . 2. Calculate the error on the energy . 3. Plot, using Excel, versus (i.e. on the X-axis, and along the Y-axis) and draw the

error bars (on the same graph). Do not forget to put the titles and units on the axes.

77



6. Does your data represent the theory given by Eq. 4? Look at the fit quality variable .

7. Calculate the mechanical equivalent of heat constant and its error (using the slope in

Eq. 4).

8. The world know value of is , how does your measured value compare with this

value?

6 Further questions

1. The water specific heat capacity is

(see the previous experiment). What

the values of in joules instead of calories.

2. If an object of mass falls, under the action of earth gravity, a distance of , what is its lost potential energy in .

3. In the old calorific liquid picture, as explained in the theory part, it was imagined that a calorific fluid just flows from the hot to the cold object until the they reach the same temperature. We now know that this picture is erroneous and that heat contained in a body comes from the kinetic energy of its constituents (atoms, molecules, etc). When a hot object is in contact with a cold one, how does, in this new picture, heat transfer from the hot to the cold object? (Hint: Think of what happens to the microscopic constituents of the two bodies).

78

References

[1] Data Reduction and Error Analysis for the Physical Sciences, P. Bevington & D Robinson, 3rd

ed.

[2] An Introduction to Error Analysis: The Study of Uncertainties in Physical Measurements, J. R.

Taylor, 2nd ed.

79

Appendix A: How to write a report

The student should strive to answer all the requirements of the experiment’s script in the lab

manual and present them in a comprehensive report. The following is an example of how such

a full report should be written.

80

Experimental General Physics for Engineers I

Student Name:

Student ID

Student Name:

Student ID

Experiment title: Free Fall

Date submitted:

Evaluation (For Instructor use only)

Date received: Due date:

Presentation, quality of text

and discussion

Data

Graph

Error analysis

Further questions

Others

Report Grade

81

1 Aim of the experiment Measurement of the free fall acceleration due to earth’s gravity.

2 Theory Near the earth surface the gravity force is constant. This means a free falling object will have a constant

acceleration . It can be shown that in such a situation (i.e. constant acceleration ) the vertical position

of the falling object in function of time is given by:

Where and are the position and the speed at , respectively. The above equation can be

written in the following form:

This is the equation which will be used in our analysis.

3 Equipments Two photo-light sensors connected to digital timer, holder, and a small ball.

4 Procedure 1. We made sure that the time measuring device worked properly.

2. We set a distance of nearly 20 cm between the two photo-sensors and measured the time

3. which the ball took to cover it. We did this measurement three times.

4. We repeated step 2 for the distances 30, 40, 50, 60 cm (or nearly so).

5. When measuring the nominated distances we tried to estimate the errors.

5 Data Following the procedure described above we obtained the first four columns from the following table. It

was straightforward to calculate the other columns need for the graph:

Time

20.0 0.129 0.125 0.118 0.124 161.3

29.9 0.160 0.162 0.162 0.1613 185.3

40.2 0.201 0.200 0.194 0.1983 202.7

50.1 0.236 0.246 0.233 0.238 210.5

82

60.2 0.261 0.260 0.260 0.2603 231.2

Remark: In the calculation of the last two columns ( and

), we kept 4 figures whenever a

rounding was necessary. The rounding rules in our case suggest, in fact, to keep only 3 digits. We kept 4

digits because these columns will be used in later calculations.

6 Data Analysis 1. Estimation of errors on and

a. We estimated the error on to (i.e. ). Although the smaller

graduation on the ruler we used is (which could be quoted as error), the situation is

that the committed errors may be slightly more than that. For example when put the ruler on

the photo-sensor, this one moves a bit. Therefore we think it’s safer to quote an error of

b. To estimate the error on we apply the standard deviation method on the last (fifth) row,

which seem to be the one which would give us the biggest error (and the safest to quote):

i. From the table the mean for this row is:

ii. √

iii.

√

√

iv. Our estimate for the error on is then (we keep only one significant digit

in the error)

2. Estimation of the error on

.

If we put

then the error propagation rules (from the manual) tells us that:

√

Using still the fifth row we get:

√

Therefore (we keep only one significant figure in the error).

3. Plot of

versus

a. Error bars: Since

is plotted on the Y axis, the error bars represent the error on

. This has

been already calculated above and found to be .

83

b. Excel graph:

versus

4. From the linear fit we have:

a. Slope

b. Intercept

c. We get the error on the slope and intercept using the Excel linest function (we notice that

the slope and intercept are the same given the Excel graph above):

466.5472 106.4914

52.58869 10.65662

0.963283 5.83735

78.70577 3

2681.871 102.224

Remark: the error on Y calculated by Excel is slightly bigger than our estimated

error . We see also that error on slope sizeable:

. This of

course will affect the precision of our measurement of .

5. Yes, our data represents the theory expressed in Eq. 3 because is close to 1, which

mean that the linear fit is a good fit. This linear relation is that what the theory predicts. (see Eq.

3).

Graph on millimetric paper: see the end of the report.

From the theory (see the experiment script) we have

. This means:

And (using error propagation)

Putting the numbers we get:

y = 466.55x + 106.49 R² = 0.9633

160

170

180

190

200

210

220

230

0.122 0.142 0.162 0.182 0.202 0.222 0.242

t_m (s)

dx/

t_m

(cm

/s)

84

So (we keep only one significant digit in the errors)

Therefore our result for is:

To compare our measurement to the known value of

, we use the

method (we assumed an error of

on the world value):

The value of is less than 3, so we conclude that our measurement is compatible with the

world value. This has been said we note that our measurement is not a good one: the error is

quite big. We can see that the error

represents

of the measured value

which a

sign that our measurement is not that precise. This also shows us that to reproduce the world

values precision of 0.1% (that is

) one needs quite a sophisticated experiment.

7 Conclusion We have measured the free fall acceleration, due to earth gravity, and obtained a value of

. This value is compatible with the world value of

. But we notice that our

measurement is only about 10 percent precise while the worlds one has a 1 per mil precision.

8 Further questions 1. What is the value of the speed of the ball when it crosses the first sensor (hint: use

your graph)? The speed of the ball when it crosses the first sensor is (initial speed). This is given by the intercept of our linear fit (see Eq. 3 in the manual). From the fit we find

2. If you plot versus , instead of

versus , how would the graph look like?

(Make a sketch). How would you calculate the acceleration and the speed when the ball crosses the first sensor?

If we plot vs we get a parabola curve (graph). This is so because the relation between and is quadratic as can be seen in Eq. 2 in the manual. The initial speed is the derivative of the curve at . This is given by the slope of the tangent to the curve at that point ( . To calculate the acceleration from this curve we can find the speed at two distinct moments and , by finding the slope of the tangents at those points, and

85

calculate

. (we can do this for several pair of point on the curve and take the

mean). A sketch of what have been said here is attached to this report. 3. In the actual experiment, if you use a very light ball (same size), would you get the

same value of ? Why? In the real actual conditions, air resistance plays an important role. A lighter ball will experience a much higher drag force due air resistance (collisions with air molecules) and thus the net force will be much smaller than the gravity force (motive force). Therefore the acceleration will be much less than (as the acceleration is proportional to the net force). 4. We neglected the air resistance exerted on the ball. Does this cause a random error

or a systematic error? How does it affect the measured value of ? As discussed previously the air resistance always decreases the ball acceleration. If we neglect the air resistance than our results will systematically smaller than the true value, here denoted (if we do many measurements, all of them will give results small than the true value). Therefore this a systematic error which we need to estimate (i.e. estimate the air resistance) and correct accordingly our result. Alternatively we can put ourselves in a situation where the air resistance is zero or really negligible ( do experiment under vacuum or use maybe a steel, dense, sphere) 5. This experiment is quite simple and seems to be accessible even to people many

centuries ago. Which measurement in the experiment do you think was difficult to achieve with any acceptable accuracy at that time (i.e. centuries ago)? Can you think of a way to reduce this difficulty? (Galileo did!).

Obviously of the two kinds of measurement we did, the time measurement would be the most difficult one centuries ago (the other measurement we did was measuring the distance between the photo-gates). The time counter we used measures milliseconds. This is achieved by using an electronic device which could not exist in the far past where only mechanical (not very precise) watches existed. When can try to overcome this difficulty, as Galileo did, by using inclined planes. There the acceleration can be very much reduced.

Graphing the data on millimetric paper

(question 6)

86

Our data is summarized in the following table:

Average

Time

0.124 161.3

0.1613 185.3

0.1983 202.7

0.238 210.5

0.2603 231.2

1. Putting the Data on X-axis:

The scale on the X-axis is defined such that (0.260 – 0.124) corresponds to 14 cm .This is chosen so

that the first entry 0.124 lies at the beginning of the X-axis (on the graph paper) and the last entry 0.260

at the end of X-axis . Any entry in the time column will then be at a distance:

Applying this formula to the time column, (i.e. X) we get:

Average Time

Position on X-

axis (cm)

0.124 0

0.1613 4.925899

0.1983 9.812179

0.238 15.05503

0.2603 18

2. Putting the data on the Y axis:

Similarly, the scale on the Y-axis is defined such that (231.2– 161.3)

corresponds to 23 cm. We then

get:

Using this formula to calculate the representation of our Y data we get:

Position on Y-

87

axis (cm)

161.3 0

185.3 7.896996

202.7 13.62232

210.5 16.09335

231.2 23

3. Representing the Error bars:

We use the Y axis scale which has been defined so that (231.2– 161.3) corresponds to 23 cm. We

have then (we have already found the error bar, i.e. the error on the Y axis, to be ):

(231.2– 161.3) 23 cm

X =

The error bars will then be represented by 1.3cm on each side of the data points.

4. Drawing the fit line:

The fit line equation given by Excel is y = 466.55x + 106.49. to draw this equation we need 2 points

(x1, y1) and (x2, y2). We take x1 = 0.124 and x2 = 0.260 (note that these are the smallest and biggest

values of time in our data) and calculate the corresponding y1 and y2:

y1= 466.55 (0.124) + 106.49 = 164

y2= 466.55 (0.260) + 106.49 = 228

therefore (x1, y1) = (0.124, 164) and (x2, y2) = (0.260, 228). We need to calculate how these 2 points will

be represented on the graph paper. We have already calculated those of x1 and x2 (we chose x1 and x2

to coincide with the data), we need then to calculate just those of y1 and y2. To do so we use the Y

scaled defined above:

we get:

Y1=

Y1=

We draw the two points and joint them by straight line, which will be the best fit (see the graph paper).

88

89

90

Appendix B: How does a Vernier caliper work?

1 Construction of the Vernier caliper The Vernier caliper, Fig. 1, has two scales: the main scale and the Vernier scale (which is

movable). It is more precise than a simple ruler (see paragraph 5 below). The trick is in the fact

that the Vernier scale is graduated in such a way that 10 of its graduation cover only 9

graduations of the main scale (this is true only for the caliper shown in Fig. 1, other types have

different segmentation. See the note by the end of this document). In other words, if a

graduation of the main scale covers 1 mm then a graduation from the Vernier scale covers only

mm (i.e. less than a millimeter). Because of this, when measuring nothing, the only

graduations on the Vernier scale which align with another on the main scale are the first and

the last one (i.e. 0 and 10).

Figure 7: Vernier caliper with its two scales

2 How to read the Vernier Caliper Assume we are measuring a distance as shown on Fig.2. One can clearly see on the main (fix)

scale that this distance is between 1mm and 2mm. So, from the main scale, equals at least 1

mm. We can then use the Vernier scale to read the first decimal as follow. We find the first

graduation on the Vernier scale that coincides with one on the main scale. We can see that the

3rd graduation does so (see Fig. 2). The Vernier caliper reading rule says that the required extra

distance is to be read as

mm (i.e. we divide the number of graduations on the Vernier scale,

here 3, by 10). The total distance is then mm mm. (Note that this is true

only when we use mm as unit).

91

Figure 8: Measuring an length with the Vernier

3 Rational Explanation From Fig. 3 below we can see that where is easily read, from the main scale,

to be 1mm. To read we need to use both the main and the Vernier scale. We can see from

the figure that , where should be read from the main scale and from the

Vernier scale. It’s clear that mm and

mm (remember that a graduation o the

Vernier scale is

mm). Therefore

mm and finally

mm (as found previously).

Figure 9: illustration of the way the Vernier works

4 Important Note Not all Vernier calipers are constructed as described in paragraph 1. Perhaps the main

difference is in how the Vernier scale is related to the main scale. But the same rules and

explanations given above can be applied for the other calipers as well. Especially, one gets a

measurement in three steps:

1. One gets a rough value by reading the main scale (in mm).

2. One then gets the precision reading by finding the coincidence graduation, reading its

position on the Vernier scale and dividing the result by 10 (in mm).

3. Add together the results of the two previous steps.

92

5 How to estimate the error We can estimate the error committed when using a Vernier caliper in the following way. We

notice first that the error comes in because we are unable to locate exactly the first graduation

on the Vernier scale that coincides with one on the main scale (see Fig. 4). Things being so we

can fairly assume that, in most cases, we would miss the coincidence graduation by at most one

graduation (on the Vernier scale). But we know that one graduation reading on the Vernier

scale corresponds to 0.1 graduation on the main scale (remember that we have to divide by 10

to translate a reading on the Vernier scale to a reading on the main scale (see paragraph 2).

Therefore the quoted error should be 0.1 x the smallest graduation on the main scale. Compare

this error with the one which we should quote if we have used the main scale only (i.e. a

normal ruler). This should be the smallest graduation. The Vernier caliper improves thus the

precision by 10 times.

figure 4: A case where it's difficult to pick up only one coincidence graduation

6 For practice http://www.members.shaw.ca/ron.blond/index.html

93

Appendix C: Least-square fit

We will use the method of Least Square to fit our data to a straight line. The aim is to find the

line that represents best our measurements. For this we have to minimize the following

quantity:

∑ *

+

∑ *

+

(12)

where and

are our measured data and uncertainties which correspond to the

independent measured variables . is the predicted (theoretical) value at

. (We

will drop in what follows the scripts and ). As one can see the , pronounced chi-

square, quantity is the sum of the square of the distances, called residuals (see Fig. 1), between

the measured data points and the theoretical straight line . Therefore

minimizing the means requiring the residuals to be as small as possible and the straight line

which satisfies this requirement is the best fit line48. The minimization is, of course, with respect

to the two free parameters, and of the theoretical line.

48

Here is a deeper reason, given by the Maximum Likelihood method, why we do minimize the . We assume first that the measurements are normally distributed (i.e. Gaussian) with Means situated on the best fit line and Standard Deviations given by the measurements error. With this assumption, the

probability for making the observed measurement is

√

[

] . This follows from

the statistic interpretation of the errors (with the Gaussian distribution as parent). The probability for making all the observed ’s is, following probability multiplication rule, the product of the individual probabilities

∏

√ {

*

+

} ,∑

[

]

- ∏

√

. The best fit line is the one that

gives the maximum probability This is the Maximum Likelihood method. But we can see that maximizing , i.e. maximizing the exponential function, is equivalent to minimizing the

94

Figure 10: Data, best fit and residuals are shown. Errors bars are equal to 5 (i.e. )

At the minimum the partial derivatives should be equal to zero:

∑[

]

∑[

]

∑[

]

∑ [

]

These two equations can be rearranged as

∑

∑

∑

∑

∑

∑

y = 9.4857x - 158.52 R² = 0.9535

405

415

425

435

445

455

465

59 61 63 65 67

Residuals

95

These are two linear simultaneous equations in the two unknown parameters and All the

other quantities are our measured (known) quantities. Finding the parameters and , by

solving these equations, means that we have found the best fit straight line . For

simplicity, we assume49 that all the quantities have the same error The solution is then

given by the following equations:

|∑

∑

∑ ∑

|

∑

∑

∑

∑

(13)

|

∑

∑ ∑

|

∑

∑

∑

(14)

| ∑

∑ ∑

| ∑

∑ (15)

Note that the solutions do not depend on the error . This is true only when the error is

common to all as we have assumed. In calculating the parameters and , one may first

accumulate the four sums ∑ , ∑

, ∑

and ∑

and then insert them in these

equations.

Now we turn to calculating the errors on the fit parameters. Since and depend on the

which have the common error , we can calculate the errors on and by using the error

propagation method50 (we assumed that have negligible errors):

∑(

)

∑[(∑

∑

)

]

∑[(∑

)

(∑

)

∑

∑

]

[ (∑

)

∑ (∑

)

∑

∑

∑

]

49

This is the case for our labs. 50

It means that the error on and come from the fact that we have errors on See also Eq.4.

96

[ (∑

)

∑ (∑

)

(∑

)

∑

]

[ (∑

)

(∑

)

∑

]

∑

* ∑

(∑

)

+

∑

∑

We have used in this derivation ∑ and the fact that ∑

∑

, ie the indexes

and are just dummy variables which we can label them as we wish.

Thus the error on (the intercept) is given by:

∑

(16)

with a similar calculation, we get the error on (the slope):

(17)

Fit Example:

Suppose we have the following measurements (this is the data used in figure 1 above):

Table 1

X Y

60 413

61 415

62 433

63 436

64 453

65 456

97

We would like to calculate the slope , the intercept and the error on them using these

measurements and the formulae developed above. First, as suggested, let us calculate the

needed sums (as we see on the table, the number of measurement is equal to 6). You can

use Excel to do these summations:

∑

∑

∑

∑

Then the determinant is (see Eq. 15):

∑

∑

The intercept is (see Eq. 13):

(∑

∑

∑

∑

)

The slope is (see Eq. 14)

( ∑

∑

∑

)

Note that the values obtained here for and are the same values as those found by Excel (see

Fig. 1). Note also that we kept the same number of significant figures as done in Excel (Fig.1) to

compare our results with the result there.

We can continue on and calculate the errors on and . Note, on the graph of Fig.1, that the

common error (this is the error bar on the y axis):

The error on the intercept is (see Eq. 16):

∑

98

Then

The error on the slope is (see Eq. 17):

Thus and

Note: The values of the errors and obtained here are not equal to the

ones obtained with Excel51 ( 1.0468 65.45) . The reason is that Excel does not use the

error stated in the error bars (i.e. ) but uses an error calculated from the spread of the

data points themselves using the following formula:

∑[ ]

(18)

Let us calculate this in our case and try to find Excel results. The following table shows the

calculation needed steps: Starting from the values of and , we calculate the values of

, then , and finally [ ] .

=

[ ]

60 413 410.622 2.378 5.655

61 415 420.108 -5.108 26.089

62 433 429.593 3.407 11.605

63 436 439.079 -3.079 9.481

64 453 448.565 4.435 19.671

65 456 458.050 -2.050 4.204

From Eq. 18 and the last column in the table above we have:

51

See the tutorial on how to calculate the errors using Excel.

99

∑[ ]

∑[ ]

We use this value in evaluating as follow:

1.095782

Which means

For we have:

∑

Then

We can see that we have obtained, with the new , the

100

Appendix D: Simple Pendulum

The simple pendulum52 equation of motion is given by (see Eq. 4 in the Simple Pendulum

experiment sheet):

(1)

We can re-write

in the following way:

Let

. Then

. Eq. 1 becomes:

(2)

Or, by separation of the variables

(3)

Integrating this last equation gives:

(4)

To determine the integration constant , we use the initial conditions. We assume that at

, the pendulum is at rest at an angle (with respect to the perpendicular). In other words:

when (5)

Putting in Eq. 4 and using the above condition ( ) we get :

Or, equivalently:

(6)

We can then re-write Eq. 4 in the following form:

52

Here we follow closely the text in ‘Advanced Calculus’ Murray R. Spiegel, Pages 340, 333 and 229.

101

√

(7)

Or

√

(8)

To fix the sign in Eq. 8 we reason as follow: When the pendulum moves from to ,

the angle is decreasing so that is negative. being positive, we have to chose the

negative sign:

√

(10)

The motion from to takes one quarter of the period

, thus:

√

∫

√

(11)

This last equation can be re-written as:

√

∫

√

√

∫

√( ) (

)

√

∫

√

where we have used

.

Changing the integration variable from to , using

, the integral becomes:

√

∫

√

(12)

102

where

is a constant. One can already see’s that for small oscillations where

the period is given by √

as was found in the Simple Pendulum experiment sheet.

The evaluation of the integral in Eq. 12 can be done as follow. The Binomial theorem, or Taylor

series, tells us that:

In our case . Thus

√

∫ [

]

√

[

]

Which is the formula given Simple Pendulum experiment sheet. In the last step we have used

the following general formula ( being positive integer):

∫

{

(see the reference of this appendix for a proof).

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