4.1.1Describe examples of oscillation. 4.1.2Define the terms displacement, amplitude, frequency, period, and phase difference. 4.1.3Define simple harmonic.

4.1.1 Describe examples of oscillation.

4.1.2 Define the terms displacement, amplitude, frequency, period, and phase difference.

4.1.3 Define simple harmonic motion (SHM) and state the defining equation as a = -2x.

4.1.4 Solve problems using the defining equation a = -2x for SHM.

4.1.5 Apply the equations as solutions to the defining equation a = -2x.

4.1.6 Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during SHM.

Topic 4: Oscillations and waves4.1 Kinematics of simple harmonic motion

Describe examples of oscillation.

Oscillations are vibrations which repeat themselves.


EXAMPLE: They can be driven externally, like a pendulum in a gravitational field.

EXAMPLE: They can be driven internally, like a mass on a spring.

x

v = 0

v = vmax

v = 0

v = 0

v = vmax

v = 0

FYI

In all oscillations v = 0 at the extremes and v = vmax in the middle of the motion.

Describe examples of oscillation.

Oscillations are vibrations which repeat themselves.


EXAMPLE: They can be very rapid vibrations such as in a plucked guitar string or a tuning fork.

Define the terms displacement, amplitude, frequency, period, and phase difference.

Consider a mass on a spring that is displaced 4 meters to the right and then released.

We call the maximum displacement x0 the amplitude. In this example x0 = 4 m.

We call the point of zero displacement the equilibrium position.

The period T (measured in s) is the time it takes for the mass to make one complete oscillation or cycle.

For this particular oscillation, the period T is about 24 seconds (per cycle).


x0

x

equili

brium


The frequency f (measured in Hz or cycles/s) is defined as how many cycles (oscillations, repetitions) occur each second.

Since period T is seconds per cycle, frequency must be 1/T.


f = 1/T relation between T and fT = 1/f

EXAMPLE: The cycle of the previous example repeated each 24 s. What are the period and the frequency of the oscillation?

SOLUTION:

The period is T = 24 s.The frequency is f = 1/T = 1/24 = 0.042 Hz


We can pull the mass to the right and then release it to begin its motion:

Or we could push it to the left and release it:The resulting motion would have the same values for T, f, and .However, the resulting motion will have a phase difference of half a cycle.


Start stretched

Start compressed

The two motions are half a cycle out of phase.x

x

PRACTICE: Two identical mass-spring systems are started in two different ways. What is their phase difference?

The phase difference is a quarter of a cycle.



Start stretched and then release

Start unstretched with a push left

x

x

PRACTICE: Two identical mass-spring systems are started in two different ways. What is their phase difference?

The phase difference is three quarters of a cycle.



Start stretched and then release

Start unstretched with a push right

x

x

Define simple harmonic motion (SHM) and state the defining equation as a = -2x.

Before we define simple harmonic motion, which is a special kind of oscillation, we have to digress for a moment and revisit uniform circular motion.

Recall that UCM consists of the motion of an object at a constant speed v0 in a circle of radius x0.

Since the velocity is always changing direction, we saw that the object had a centripetal acceleration given by a = v0

2/x0, pointing toward the center.

If we time one revolution we get the period T.

T is about 12 s.And the frequency is f = 1/T = 1/12 = 0.083 s.


x0

v0


We say that the angular speed of the object is 360 deg/12 s = 30 deg s-1.

In Topic 4 we must learn about an alternate and more natural method of measuring angles besides degrees.

They are called radians.


rad = 180° = 1/2 rev radian-degree-revolution conversions2 rad = 360° = 1 rev

EXAMPLE: Convert 30 into radians (rad) and convert 1.75 rad to degrees.

SOLUTION:

30( rad / 180° ) = 0.52 rad. 1.75 rad (180° / rad ) = 100°.


Angular speed will not be measured in degrees per second in this section. It will be measured in radians per second.



EXAMPLE: Convert the angular speed of 30 s-1 from the previous example into radians per second.

SOLUTION:

Since 30( rad / 180° ) = 0.52 rad,then 30 s-1 = 0.52 rad s-1.FYI

Angular speed is also called angular frequency.

x0

v0


Since 2 rad = 360° = 1 rev it should be clear that the angular speed is just 2/T.

And since f = 1/T it should also be clear that = 2f.


= 2/T = /t relation between , T and f = 2f


EXAMPLE: Find the angular frequency (angular speed) of the second hand on a clock.

SOLUTION:

Since the second hand turns through one circle each 60 s, it has an angular speed given by

= 2/T = 2/60 = 0.105 rad s-1.



= 2/T = /t relation between , T and f = 2f


EXAMPLE: A car rounds a 90° turn in 6.0 seconds. What was its angular speed during the turn?

SOLUTION:

Since needs radians we begin by converting : = 90°( rad / 180° ) = 1.57 rad.Now we use = /t = 1.57/6 = 0.26 rad s-1.



PRACTICE: Find the angular frequency of the minute hand of a clock, and the rotation of the earth in one day.The minute hand takes 1 hour to go around one time. Thus = 2/T = 2/3600 s = 0.0017 rad s-1.The earth takes 24 h for each revolution so that = 2/T = ( 2 / 24 h )( 1 h / 3600 s ) = 0.000073 rad s-1.This small angular speed is why we can’t really feel the earth as it spins.



PRACTICE: This is an old IB question. An object is traveling at speed v0 in a circle of radius x0. The period of the object’s motion is T.(a) Find the speed v0 in terms of x0 and T.Since the object travels a distance of one circumference in one period v0 = distance / time v0 = circumference / period v0 = 2x0/T.(b) Show that v0 = x0.Since v0 = 2x0/T and = 2/T we have v0 = 2x0/T v0 = x02/T v0 = x0(2/T) v0 = x0

x0

v0



PRACTICE: This is an old IB question. An object is traveling at speed v0 in a circle of radius x0. The period of the object’s motion is T.(c) Find the centripetal acceleration aC in terms of x0 and .Since the centripetal acceleration is aC = v0

2/x0 and since v0 = x0, v0

2 = x0

22

aC = v02/x0

aC = x022/x0

aC = x02. x0

v0


You might be asking yourself how an oscillating mass- spring system might be related to uniform circular motion. The relationship is remarkable.

Consider a rotating disk that has a ball glued onto its edge. We project a strong light to produce a shadow of the ball’s motion on a screen.

Like the mass in the mass- spring system, the ball behaves the same at the arrows:


x0

x

x 0

x 0 x 0

x0

x0

x0

x0x0 x0

x0

x0

x0

-x0 x00

x


Note that the shadow is the x-coordinate of the ball.

Thus the equation of the shadow is x = x0 cos .

Therefore the equation of the shadow’s x-coordinate is

x = x0 cos t.

If we know , and if we know t, we can calculate x.

x0

v0x 0

v 0


FYI

From = /t we get = t.

x-x0 x00

x

x = x0 cos


At precisely the same instant we can find the equation for the shadow of v.

Create a velocity triangle.Working from the displacement triangle we can determine the angles in the velocity triangle.

The x-component of the velocity is opposite the theta, so we use sine: v = -v0 sin .

But since = t we get our final equation:

v = -v0 sin t.

Why is our v negative?


x 0

v0

90-

90-


Remember the acceleration of the mass in UCM?We found recently that aC = x02.

And we know that it points to the center.

The x-component of the acceleration is adjacent to the theta, so we use cosine: a = -aC cos .

But since = t we get our final equation and aC = x02

we have

a = -x02 cos t.

Why is our a negative?Since x = x0 cos t we can just write a = -2x.


x 0

v0

aC


Let’s put all of our equations in a box-there are quite a few!

We say a particle is undergoing simple harmonic motion (SHM) if it’s acceleration is of the form a = -2x.

Data Booklet has highlighted formulas.


x = x0 cos t Set 1 - equations of simple harmonic motionv = -v0 sin t

a = -x02 cos t

a = -2x x0 is the maximum displacementv0 is the maximum speed

= 2/T = 2f

This equation set works only for a mass which begins at x = +x0 and is released from rest at t = 0 s.

v0 = x0

x0

v0

x


Without deriving the other set in the Physics Data Booklet, here they are:

If you are interested, this last set is derived from a clockwise-rotation beginning as shown:

Data Booklet has highlighted formulas.


x0 v0

x

x = x0 sin t Set 2 - equations of simple harmonic motionv = v0 cos t

a = -x02 sin t

a = -2x x0 is the maximum displacementv0 is the maximum speed

= 2/T = 2f

This equation set works only for a mass which begins at x = 0 and is given a positive velocity v0 at t = 0 s.

v0 = x0


We need one more formula before we can practice.Set 1: x = x0 cos t, v = -v0 sin t and v0 = x0:

Begin by squaring each equation from Set 1: v2 = (-v0 sin t)2 = v0

2 sin2

t.

v02 = x0

22, and x2 = x02 cos2

t.Then v2 = v0

2 sin2

t becomes v2 = x0

22 sin2 t.

sin2 t + cos2 t = 1 yields sin2 t = 1 - cos2 tso that v2 = x0

22(1 - cos2 t) or

v2 = 2(x02 - x0

2 cos2 t)

Then v2 = 2(x02 - x2), which becomes


v = (x02 - x2) relation between x and v

Apply the equations as solutions to the defining equation a = -2x.


EXAMPLE: A spring having a spring constant of 125 n m-1 is attached to a 5.0-kg mass, stretched +4.0 m as shown, and then released from rest.

(a) Using Hooke’s law, show that the mass-spring system undergoes SHM with 2 = k/m.

SOLUTION:

From Hooke’s law F = -kx.From Newton’s second law F = ma.Thus ma = -kx or a = -(k/m)x. The result of a = -(k/m)x is of the form a = -2x where 2 = k/m.

Therefore, the mass-spring system is in SHM.

x




(b) Find the angular frequency, frequency and period of the oscillation.

SOLUTION:

Since 2 = k/m = 125/5 = 25 then = 5 rad s-1.

Since = 2f, then f = /2 = 5/2 = 0.80 Hz.Since = 2/T, then T = 2/ = 2/5 = 1.3 s.

x




(c) Show that the position and velocity of the mass at any time t is given by x = 4 cos 5t and that v = -20 sin 5t.

SOLUTION:

Note: 2 = k/m = 125/5 = 25 so = 5 rad s-1.

Note: x0 = 4 m, and v0 = xo = 4(5) = 20 m s-1.

At t = 0 s x = +x0 and v = 0, so use Set 1:

x = x0 cos t and v = -v0 sin t

x = 4 cos 5t and v = -20 sin 5t.

x




(d) Find the position, the velocity, and the acceleration of the mass at t = 0.75 s. Then find the maximum kinetic energy of the system.

SOLUTION:

x = 4 cos 5t = 4 cos 5(.75) = 4 cos 3.75 = -3.3 m.

v = -20 sin 5t = -20 sin 5(.75) = +11 m s-1.

a = -2x = -52(-3.3) = 82.5 m s-2.

EK,max = (1/2)mvmax2

= (1/2)mv02 = (1/2)(5)(202) = 1000 J.

x

Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during SHM.


EXAMPLE: The displacement x vs. time t for a 2.5-kg mass is shown in the sinusoidal graph.

(a) Find the period, the angular velocity, and the frequency of the motion.

SOLUTION:

The period is the time for one complete cycle. It is T = 6.010-3 s.

= 2/T = 2/0.006 = 1000 rad s-1. [1047]

f = 1/T = 1/0.006 = 170 Hz.



EXAMPLE: The displacement x vs. time t for a 2.5-kg mass is shown in the sinusoidal graph.

(b) Find the velocity and acceleration of the mass at t = 3.4 ms.

SOLUTION:

From the graph x = -0.810-3 m at t = 3.4 ms.

From the graph x0 = 2.010-3 m. Thus

v = (x02 - x2) = -1047 (0.0022 - 0.00082)

= -1.9 m s-1.

Finally a = -2x = -(10472)(-0.0008) = +880 m s-2.

Why is v negative?Because the slope is!



PRACTICE: The displacement x vs. time t for a 2.5-kg mass is shown in the sinusoidal graph. (a) Find the period and the angular frequency of the mass. Then find the maximum velocity.The period is the time for one complete cycle. It is T = 6.0 s. = 2/T = 2/6 = 1.0 rad s-1. [1.047]v0 = x0 = 1.4(1.047) = 1.5 m s-1.



PRACTICE: The displacement x vs. time t for a 2.5-kg mass is shown in the sinusoidal graph. (b) Find the force acting on the mass at t = 3 s. Then find it’s velocity at that instant.Use F = ma where m = 2.5 kg. At t = 3 s we see that x = -1.4 m.Then a = -2x = -(1.0472)(-1.4) = 1.5 m s-2.Then F = ma = 2.5(1.5) = 3.8 n.Because the slope is zero, so is the velocity.

4.1.1Describe examples of oscillation. 4.1.2Define the terms displacement, amplitude, frequency, period, and phase difference. 4.1.3Define simple harmonic.

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period t

phase difference

oscillations v

terms displacement

simple harmonic motion

frequency f

cycles oscillations

maximum displacement