Phys 101 learning object lo6 standing waves vivian tsang

Phys 101 Learning Object LO6 March 7 2015Standing Sound Waves by Vivian Tsang 14153143

Let’s Learn the Basics• Standing waves occur when two sinusoidal waves

with the same wavelength, frequency, and amplitude travelling in opposite directions cross paths

• This can be illustrated by the following to equations for the waves:▫DR=Asin(kx-wt) *w= fundamental frequency▫DL=Asin(kx+wt) *A= amplitude

• This leads to the equation:▫D(x, t)=(2Asin(2pi/lambda)x)*(cos(wt))

◦ *k= 2pi/lambda

Nodes=zero displacement

•At a node, sin((2pi/lambda)x)=0

•Therefore, nodes occur at every interval of lambda/2

node nodenode

Lambda/2

Antinodes=max displacement

•At an antinode, sin((2pi/lambda)x)=+/-1

•Therefore, antinodes occur at every interval of lambda/2

antinode

antinode

Lambda/2

Nodes and Antinodes

•The distance between nodes and antinodes are lambda/4antinode

antinode

nodenode

node

Lambda/4

For a string fixed at both ends…• Solving for the

frequency, we can see that the fundamental frequency occurs when n=1.

• Allowed frequencies are also called harmonic sequences

• Therefore, the fundamental frequency is also called the fundamental harmonic

Allowed Frequencies• For a string fixed at both ends, the allowed frequencies include integer multiples of the fundamental harmonic•As we increase the integer n=2, we get the second harmonic•As we increase the integer n=3, we get the third harmonic

Harmonics• As you’ve noticed from the equations,

the “nth” harmonic is just an “n” multiple of the fundamental harmonic

• Therefore:▫ If the first harmonic is f1

▫ The second harmonic is f2=2f1

▫ The third harmonic is f3=3f1

▫ The fourth harmonic is f4=4f1

▫ The fifth harmonic is f5=5f1

▫ And so on …..

Practice!

QUESTION: A string vibrates at 400Hz with the pattern shown here. What is the frequency of the fourth harmonic?

?

Answer• Since we are

given the frequency of the third harmonic:

• 1) divide this frequency f3 by three to find the fundamental harmonic f1

• 2) multiply f1 by 4 to get the fourth harmonic f4!

Getting Harder! Standing Sound Waves•Longitudinal standing sound waves can

travel in the air in a tube or pipe

•At a closed end, there is a node because air molecules cannot move

•At an open end, there is an antinode because the air molecules are free to move

Basic Ideas…• Notice! the

allowed frequencies for pipes open at both ends is similar to our previous calculations for allowed frequencies on a string

• Notice! the allowed frequencies for pipes open at one end and closed at one end are only odd integer multiples

Practice!

•The fundamental frequency for a clarinet with one end closed and one end opened is 300Hz. When both ends of the same pipe are opened, what is the new fundamental frequency?

Answer!• First, we solve for

the unknown ratio of v/L for the fundamental frequency fp1of the pipe with a closed end

• Next, we substitute this ratio into the equation fp2 to find the fundamental frequency of the pipe when the two ends are open

Harder Problem!• An alphorn is a tube opened

at one end and closed at the other.

• This alphorn is 4m in length.• QUESTION: By what length

would you have to change the alphorn’s length if you want to increase its fundamental frequency by 200Hz?

• What will be the new fundamental frequency?

Hints• This is a harder problem so let’s break it down:• Given:

▫ V=the speed of sound in air which is about 343m/s▫ L= 4m▫ Since the pipe is open at one end and closed at the other,

we would use the formula f=v/4L▫ Now, you can solve for f1, the fundamental frequency

• Next:▫ Use the change of frequency given (200Hz) to solve for the

change in length L=v/4 f• Finally:

▫ Add up the original fundamental frequency and the change in frequency to find the new fundamental frequency

Solution!

Thanks for watching!