phys 101 learning object lo6 standing waves vivian tsang
TRANSCRIPT
![Page 1: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/1.jpg)
Phys 101 Learning Object LO6 March 7 2015Standing Sound Waves by Vivian Tsang 14153143
![Page 2: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/2.jpg)
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
![Page 3: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/3.jpg)
Nodes=zero displacement
•At a node, sin((2pi/lambda)x)=0
•Therefore, nodes occur at every interval of lambda/2
node nodenode
Lambda/2
![Page 4: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/4.jpg)
Antinodes=max displacement
•At an antinode, sin((2pi/lambda)x)=+/-1
•Therefore, antinodes occur at every interval of lambda/2
antinode
antinode
Lambda/2
![Page 5: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/5.jpg)
Nodes and Antinodes
•The distance between nodes and antinodes are lambda/4antinode
antinode
nodenode
node
Lambda/4
![Page 6: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/6.jpg)
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
![Page 7: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/7.jpg)
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
![Page 8: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/8.jpg)
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 …..
![Page 9: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/9.jpg)
Practice!
QUESTION: A string vibrates at 400Hz with the pattern shown here. What is the frequency of the fourth harmonic?
?
![Page 10: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/10.jpg)
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!
![Page 11: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/11.jpg)
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
![Page 12: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/12.jpg)
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
![Page 13: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/13.jpg)
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?
![Page 14: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/14.jpg)
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
![Page 15: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/15.jpg)
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?
![Page 16: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/16.jpg)
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
![Page 17: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/17.jpg)
Solution!
![Page 18: Phys 101 learning object lo6 standing waves vivian tsang](https://reader035.vdocuments.us/reader035/viewer/2022062710/55b7b891bb61eb186a8b46f2/html5/thumbnails/18.jpg)
Thanks for watching!