Physics 203 College Physics I Fall 2012

1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 12 Sound – Part 2

2. Announcements • The Final Exam is next Monday: 8 – 11 AM here. • Bring a page of notes. I won’t provide equations. I will provide constants, conversion factors, and moments of inertia. • You can find solutions to the end-of-chapter problems from each homework set and solutions to all of the exams in the Course Materials section of MasteringPhysics. • Also, see my list of equations for each exam and the last several chapters there.

3. Quiz Question 1 • If a mass hanging on the end of a light-weight spring is pulled down and released, the oscillation period will depend on • A The spring stiffness constant • B The hanging mass • C The distance it is pulled down • There may be more than one answer. Enter multiple choices sequentially.

4. Quiz Question 2 • The frequency of a wave increases. If the speed of the wave remains constant, what happens to the distance between successive crests of the wave? • A The distance increases • B The distance decreases • C The distance remains the same

5. Quiz Question 3 • When a guitar string is vibrating in its fundamental mode, the wavelength of the vibration is • A the length of the guitar string. • B half the length of the guitar string. • C twice the length of the guitar string. • D independent of the length of the guitar string.

6. Quiz Question 4 • A standing wave is created in an organ pipe which is closed on one end and open on the other. Which is true? • A There is a pressure node at both ends. • B There is a displacement node at both ends • C There is a pressure node at the closed end and a displacement node at the open end. • D There is a pressure node at the open end and a displacement node at the closed end.

7. Quiz Question 5 • If I am listening to a band at a concert, and move twice as far from the speakers, the loudness will • A be a few decibels less than before. • B be about 100 decibels less than before. • C be about half as many decibels as before. • D be about ¼ as many decibels as before.

8. Quiz Question 6 • The Doppler Effect on the frequency of a sound is • where vsnd, vobs, and vsrc are the velocities of the sound, the observer, and the source, if the sign conventions are chosen appropriately. The correct choice when the source and observer are approaching each other is • A) vobs > 0, vsrc > 0 B) vobs < 0, vsrc > 0 • C) vobs < 0, vsrc < 0 D) vobs> 0, vsrc < 0

9. Doppler Effect • If you are moving toward the source of a sound, you pass through the wave crests more rapidly than if you were standing still.

10. Doppler Shift • If you are moving toward the source of a sound, you pass through the wave crests more rapidly than if you were standing still. • The relative speed of the sound crests and you would be v’ = vsnd + vobs, so the frequency is • f’ = v’/l = (vsnd + vobs)/l with l = vsnd/f • so that • vsnd+ vobsvsnd • vobsvobs ( ) ( ) f ’ = f = + 1 f

11. Doppler Effect • If the source is moving toward you and you are still, the relative speed of sound is unchanged, but the wavelength is.

12. Doppler Effect • The fire truck emits sound-wave crests with period T = 1/f. The crest moves ahead a length l= vsndT in this time. • But the fire truck is moving too, so it is a distance D = vsrc T closer to the previous crest when it emits the next crest. • Since T = 1/f = l/vsnd, D = lvsrc/vsnd. • The wavelength is then • vsrc • vsnd ( ) l’ = l-D = l 1 –

13. Doppler Effect ( ) • Equivalently,vsnd – vsrc • vsnd • To get the frequency, use l= vsnd/f, l’ = vsnd/f’ and taking the reciprocal of both sides gives • vsnd • vsnd – vsrc • If both the source and observer are moving, combine the two expressions: l’ = l ( ) f’ = f.

14. Guitar Sound • Low A on a slightly-out-of-tune open guitar string • (Low A should be 110 Hz, not 108 Hz.) 324 E3 A3 A4 C5 E5 G5 A5 B5 216 432 540 648 756 864 972 A2 108

15. Wind Instruments • Wind Instruments produce sounds using a vibrating column of air. • For example, consider this tube, open on the ends.

16. Wind Instruments • Air can vibrate back and forth, but at the ends, the pressure must be the same as it is outside the tube. • The air can blow in and out freely, keeping the pressure fixed. So there is no pressure variation at the ends.

17. Wind Instruments • Drawing a red line for the difference between the atmospheric pressure and pressure in the pipe, the wave’s maxima and minima would look like this: L antinode node node

18. Wind Instruments • Then the frequencies produced by the open tube are • fN = Nv/2L = Nf1 L antinode node node

19. Wind Instruments • The open tube produces frequencies f1, 2f1, 3f1, 4f1, … • These are the fundamental, 2nd harmonic, 3rd harmonic, … • This is just as for a vibrating string. • f = v/2L = f1 0 f1 2f1 3f1 4f1 5f1 Df Df Df Df Df

20. Wind Instruments • If the tube is closed on one end, the frequencies produced are different. Then the pressure variation is a maximum at the closed end, where the air cannot move. L anti- node node

21. Wind Instruments • The fundamental mode is produced with ¼ wavelength in the tube. • This means that l1 = 4L, f1 = v/l1 = v/4L. L anti- node node

22. Wind Instruments • The next harmonic occurs at the next higher frequency with an antinode on the left: • Then 3l/4 = L, f = v/l = 3v/4L = 3f1. L anti- node anti- node node

23. Wind Instruments • Since f = 3f1, this is called the thirdharmonic. • There is no second harmonic in a tube open on one end and closed on the other. L anti- node anti- node node

24. Wind Instruments • In general, for a tube closed on one end, • fN = Nv/4L = Nf1 • with N = 1, 3, 5, 7, 9, … L anti- node anti- node node

25. Wind Instruments • Notice that the difference between successive harmonics still corresponds to a half wavelength, as for an open tube: • D f = v/2L ( = f3 – f1, …) L anti- node anti- node node

26. Wind Instruments • The closed tube produces frequencies f1, 3f1, 5f1, 7f1, … • These are the fundamental, 3rd harmonic, 5th harmonic, … • The even harmonics are all missing. • f = v/2L = 2f1 0 f1 3f1 5f1 7f1 9f1 Df Df Df Df Df/2

27. Wind Instruments • Example: Didgeridoo – the first two harmonics produced are (analyzed using my laptop) • f1 = 74 Hzf3= 214 Hz • Compare 3 f1 = 222 Hz T1 = 1/f1 = 0.0113 s T3 = 1/f3 = 0.00467 s

28. Actual Frequencies Produced • The actual frequencies produced by an instrument are a superposition of the possible harmonics, with one of them usually being dominant. • Didgeridoo… (± means slightly sharp/flat) 74 D+ 148 D+ 222 A+ 369 F# - 516 C- • 295D+ 443 A+ Frequency (Hz)

29. Wind Instruments • We can find the speed of sound using the frequencies and length of the tube. • Df = f3 - f1 = 140 Hz = v/2L • L = 1.20 m, giving v = 2LDf = 341 m/s. • This is very close to the expected result (343 m/s). This works for either an open or closed tube!

30. Vocal Harmonics • Higher harmonics can also be produced strongly from the voice, a type of wind instrument. This is used, for example, in Tuvan throat singing to produce several notes simultaneously, or to generate very high pitched sounds – up to the 16th - 18thharmonic. • There are no instruments or whistling here, just one person singing. Mergen Mongush, on TUVA, Voices from the Center of Asia – Smithsonian/Folkways Records