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