Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 12 Sound – Part 2

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel 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

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel Quiz Question 6 The Doppler Effect on the frequency of a sound is where v snd, v obs, and v src 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) v obs > 0, v src > 0 B) v obs 0 C) v obs 0, v src < 0

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel 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’ = v snd + v obs, so the frequency is f’ = v’/  = (v snd + v obs )/  with  = v snd /f so that v snd + v obs v snd v obs v obs f ’ = f = + 1 f ( )

Physics 203 – College Physics I Department of Physics – The Citadel Doppler Effect If the source is moving toward you and you are still, the relative speed of sound is unchanged, but the wavelength is.

Physics 203 – College Physics I Department of Physics – The Citadel Doppler Effect The fire truck emits sound-wave crests with period T = 1/f. The crest moves ahead a length = v snd T in this time. But the fire truck is moving too, so it is a distance D = v src T closer to the previous crest when it emits the next crest. Since T = 1/f = /v snd, D = v src /v snd. The wavelength is then v src v snd ’ =  D  1 – ( )

Physics 203 – College Physics I Department of Physics – The Citadel Doppler Effect Equivalently, v snd – v src v snd To get the frequency, use = v snd /f, ’ = v snd /f’ and taking the reciprocal of both sides gives v snd v snd – v src If both the source and observer are moving, combine the two expressions: ’  ( ) f’ = f. ( )

Physics 203 – College Physics I Department of Physics – The Citadel Guitar Sound Low A on a slightly-out-of-tune open guitar string (Low A should be 110 Hz, not 108 Hz.) A2 A3 A4 E3 C5 E5 G5 A5 B5

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments Wind Instruments produce sounds using a vibrating column of air. For example, consider this tube, open on the ends.

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel antinode 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 node

Physics 203 – College Physics I Department of Physics – The Citadel antinode Wind Instruments Then the frequencies produced by the open tube are f N = Nv/2L = Nf 1 L node

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments The open tube produces frequencies f 1, 2f 1, 3f 1, 4f 1, … These are the fundamental, 2 nd harmonic, 3 rd harmonic, … This is just as for a vibrating string. 0 f 1 2f 1 3f 1 4f 1 5f 1 ff ff ff ff  f = v/2L = f 1 ff

Physics 203 – College Physics I Department of Physics – The Citadel 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 node anti- node

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments The fundamental mode is produced with ¼ wavelength in the tube. This means that  1 = 4L, f 1 = v/ 1 = v/4L. L node anti- node

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments The next harmonic occurs at the next higher frequency with an antinode on the left: Then  3  = L, f = v/ = 3v/4L = 3f 1. L node anti- node anti- node

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments Since f = 3f 1, this is called the third harmonic. There is no second harmonic in a tube open on one end and closed on the other. L node anti- node anti- node

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments In general, for a tube closed on one end, f N = Nv/4L = Nf 1 with N = 1, 3, 5, 7, 9, … L node anti- node anti- node

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments Notice that the difference between successive harmonics still corresponds to a half wavelength, as for an open tube:  f = v/2L ( = f 3 – f 1, …) L node anti- node anti- node

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments The closed tube produces frequencies f 1, 3f 1, 5f 1, 7f 1, … These are the fundamental, 3 rd harmonic, 5 th harmonic, … The even harmonics are all missing. 0 f 1 3f 1 5f 1 7f 1 9f 1 ff ff ff ff  f = v/2L = 2f 1  f/2

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments Example: Didgeridoo – the first two harmonics produced are (analyzed using my laptop) f 1 = 74 Hz f 3 = 214 Hz Compare 3 f 1 = 222 Hz T 1 = 1/f 1 = s T 3 = 1/f 3 = s

Physics 203 – College Physics I Department of Physics – The Citadel 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) Frequency (Hz) 148 D+ 222 A+ 369 F# C- 74 D+ 295 D+ 443 A+

Physics 203 – College Physics I Department of Physics – The Citadel Wind Instruments We can find the speed of sound using the frequencies and length of the tube.  f = f 3 - f 1 = 140 Hz = v/2L L = 1.20 m, giving v = 2L  f = 341 m/s. This is very close to the expected result (343 m/s). This works for either an open or closed tube!

Physics 203 – College Physics I Department of Physics – The Citadel 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 16 th - 18 th harmonic. There are no instruments or whistling here, just one person singing. Mergen Mongush, on TUVA, Voices from the Center of Asia – Smithsonian/Folkways Records