Waves Standing Waves and Sound

Transcription

1 Waves Standing Waves and Sound Lana Sheridan De Anza College May 24, 2018

2 Last time sound

3 Overview interference and sound standing waves and sound musical instruments

4 Speed of Sound waves v = B ρ Compare this expression to the speed of a pulse on a string. T v = µ Both of these expressions can be thought of as: elastic quantity v = inertial quantity These expressions are the same in spirit, but the precise quantities are the ones that represent elasticity and inertia in each case.

5 Speed of Sound in Air For air the adiabatic bulk modulus B = Pa and at 20 C. ρ = kg/m 3 This gives a speed of sound in air at 20 C of v = 343 m/s This is approximately 1/3 km/s or 1/5 mi/s.

6 Speed of Sound in Air The speed of sound in air at 20 C v = 343 m/s Since the density of air vary with temperature, the speed of sound varies also. For temperatures near room temperature: v = (331 m/s) where T Cel is the temperature in Celsius. 1 + T Cel The bulk modulus depends on the pressure.

7 Pressure Waves P(x, t) = P max sin(kx ωt) where P max = B s max k It is easier to express the amplitude in terms of the wave speed, since it is usually easier to look up the wave speed than the bulk modulus: P max = (ρv 2 ) s max ω v Then P max = ρvωs max

8 Sound Waves Displacement: s(x, t) = s max cos(kx ωt) Pressure: where P(x, t) = P max sin(kx ωt) P max = B s max k = ρvωs max

9 ns that they. Interference 17-7a, path of Sound Waves e wave Imagine from two point sources of sinusoidal sound waves that emit not identical be insignals: same amplitude, wavelength, and phase. nds on their P 1 (x, t) = P 2 (x, t) = P max sin(kx ωt) recall (from length.thus, P L 1 (17-20) S 1 L 2 (17-21) S 2 The interference at P depends on the difference in the path lengths to reach P. (a) ger multiple The sound will be louder at different points, depending on the difference in the path lengths that the sound waves take. (17-22) P

10 owing Young, 3 we consider a line source of cylindrical light waves of wavelength λ t inge Interference on a planar screenof that Sound is parallel Waves to the line source at closest distance large compa The screen has two narrow line slits, parallel to the line source and separated by dista ocatedsymmetricallyaboutthelineonthescreenclosesttothesource. Interference pattern from two sources, with equal wavelength andbeyond ar screen in phase: is a cylindrical screen of radius R λ on which the intensity of the ligh rved. e first to consider scalar waves (i.e., scalardiffraction theory), with amplitude Ae ach of the two slits, where A is a complex number, ω = kc =2πc/λ and c is the speed in vacuum 1 Thomas The intensity Young, On of the a wave Nature at of some Light point and is Colours, proportional Lecture to 39, the Course absolute of squ s amplitude Lecturesthere. on Natural We suppose Philosophy theand units Mechanical of A are Arts such (London, that the 1897) (time-average) pow

11 we Interference recall (from of Sound Waves avelength.thus, L 1 P (17-20) S 1 L 2 (17-21) S 2 The interference at P depends on the difference in the path lengths to reach P. (a) nteger multiple This is because the path difference will correspond to a phase offset of the arriving waves at P: ce). (17-22) P(x, t) = P 1 + P 2 = P max (sin(kl If the 1 difference ωt) + sin(kl is equal 2 ωt)) to, [ ( )] ( (17-23) say, k(l2 2.0 λ, then L 1 ) the waves k(l2 + L 1 ) = 2P max cos arrive exactly in phase. sin 2 This 2 a is equal to 2l, is how transverse waves new amplitude at point P (Fig. would look. P ) ωt

12 Interference of Sound Waves The new amplitude could be written as: ( ) π(l2 L 1 ) 2P max cos λ When L 2 L 1 = nλ and n = 0, 1, 2,... the sound from the two speakers is loudest (a maximum). When L 2 L 1 = (2n+1)λ 2 and n = 0, 1, 2,... the sound from the two speakers is cancelled out (a minimum).

13 Example 18.1 AM by the Same Source Two identical loudspeakers placed 3.00 m apart are driven by the t are driven same oscillator. by the same Aoscillator listener is (Fig. originally 18.5). A atlistener point O, is originally located 8.00 at m line connecting from the center the two ofspeakers. the line connecting The listener then two moves speakers. to point The P, m O, and listener she experiences then movesthe to first point minimum P, which in sound is a perpendicular intensity. What distance m from O, and she experiences the first minimum in sound intensity. What is the frequency of the oscillator? enters a nt paths xample, plit and ving the sition of splitting o Figure 3.00 m 1.15 m r m r m m P Figure 18.5 (Example 18.1) Two identical loudspeakers emit sound waves to a listener at P. O 1.85 m o separate 0 Serway sources & Jewett, combine, page we 537 apply the waves in interference analysis

14 Example 18.1 enters a nt paths xample, plit and ving the sition of splitting o Figure 3.00 m 1.15 m 8.00 m 8.00 m o separate sources combine, we apply the waves in interference analysis continued r 1 r m P Figure 18.5 (Example 18.1) Two identical loudspeakers emit P is first sound minimum. waves to a listener at P. O 1.85 m

15 Example 18.1 enters a nt paths xample, plit and ving the sition of splitting o Figure 3.00 m 1.15 m 8.00 m 8.00 m o separate Thatsources means combine, r 2 r 1 = we λ apply the waves in interference analysis 2. If we find λ, we can find f, since we know the speed of sound. continued r 1 r m P Figure 18.5 (Example 18.1) Two identical loudspeakers emit P is first sound minimum. waves to a listener at P. λ = 2( ) = 0.26 m O 1.85 m f = v λ = 343 m/s 0.26 m = 1.3 khz

16 Standing Sound Waves in air columns Standing sound waves can be set up in hollow tubes. This is the idea behind how pipe organs, clarinets, didgeridoos, etc. work. Displacement fluctuation: s(x, t) = [2s max cos(kx)] cos(ωt) For a tube with a closed end, the closed end forms a displacement node. This is logical because the air cannot move past the sealed end. (But that does also mean that the closed end is a pressure antinode. Here we will speak about sound waves in terms of displacement.)

17 Standing Sound Waves in air columns The open end or ends of a tube are approximately 1 displacement antinode. This can be thought of as being because the pressure outside the tube is atmospheric pressure, P 0, so the open end is a pressure node, therefore a displacement antinode. The waves are partially reflected from the open end because the air outside the tube can expand in 3-dimensions, rather than just one, so it behaves very differently than the air in the tube. It is effectively a change of medium. 1 This is not exactly true. Actually the antinode is located just beyond the end of the tube. For this course, we will say the the open end is an antinode.

18 In a pipe open at both ends, the ends are displacement antinodes and the harmonic series contains Standing Sound Waves in air columns all integer multiples of the fundamental. Two open ends open end is a displacement antinode and the closed end is a node. The harmonic series contains only odd integer multiples of the fundamental. One closed end elem gitu ope umn L L First harmonic A N A A N First harmonic l1 2L f 1 v v l1 2L l1 4L f 1 v v l1 4L Second harmonic A N A N A A N A N Third harmonic l2 L f 2 v 2f L l L 3 f 3 3v 3f 4L 1 A A A A A Third harmonic N N N 3 2 l 3 L f 3 3v 3f 2L 1 a b 1 Figure from Serway & Jewett, page 547. N A A N 5 4 l 5 L f 5 5v 5f 4L 1 N Fifth harmonic

19 Standing Sound Waves in air columns For double open ended tubes: The wavelengths of the normal modes are given by the constraint cos(0) = cos(kl) = 1: λ n = 2L n where n is a positive natural number (1, 2, 3...). The natural frequencies: f n = nv 2L = n f 1 where n is a positive natural number. The fundamental frequency, also called the first harmonic is the lowest frequency sound produced in the column. It is f 1 = v 2L

20 Standing Sound Waves in air columns For tubes with one closed end: fundamental frequency f 1 = v 4L The wavelengths of the normal modes are given by the constraint cos(0) = 0, cos(kl) = 1: λ 2n+1 = 4L (2n + 1) where n is a natural number including zero (0, 1, 2, 3...). The natural frequencies: f 2n+1 = (2n + 1)v 4L = (2n + 1)f 1 where n is a natural number including zero (0, 1, 2, 3...).

21 Musical Instruments Didgeridoo: Longer didgeridoos have lower pitch, but tubes that flare outward have higher pitches this can also change the spacing of the resonant frequencies. 1 Matt Roberts via Getty Images.

22 Musical Instruments, Pipe Organ The longest pipes made for organs are open-ended 64-foot stops (tube is effectively 64 feet+ long). There are two of them in the world. The fundamental frequency associated with such a pipe is 8 Hz. 32 stops give 16 Hz sound, 16 stops give 32 Hz, 8 stops give 64 Hz, etc. 1 Picture of Sydney Town Hall Grand Organ from Wikipedia, user Jason7825.

23 Musical Instruments 458 CHAPTER 17 WAVES II Bass saxophone Baritone saxophone Tenor saxophone Alto saxophone Soprano saxophone More gen open ends co where n is ca write the reso f v Figure 17 Violin standing sou Viola open end. As Cello the closed en Bass having a wav pattern requ In general, larger instruments can create lower tones, whether A B C D E F G A B C D E F G A B C D E F G A B C D E F G A B C D E F G A B C D E F G A B C D E F G A B C so on. string instruments or tube instruments. More ge Fig The saxophone and violin 1 Halliday, Resnick, only one open end families, Walker, showing 9th ed, the page relations 458. between in-

24 Summary standing waves and sound musical instruments

25 Announcements and HW Collected Homework, due Tuesday, May 29. Drop Deadline Friday, June 1. 3rd Test Friday, June 1. Quiz tomorrow. No Class on Monday May 28. (Memorial day) Homework Serway & Jewett: Ch 18, onward from page 555. OQs: 3, 5; CQs: 1, 3; Probs: 37, 39, 49, 51, 53