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# Lecture 09 interference for sound waves. beats. doppler effect

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Lecture 09 interference for sound waves. beats. doppler effect

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### Lecture 09 interference for sound waves. beats. doppler effect

1. 1. Lecture 9 Interference for sound waves. Beats. Doppler effect.
2. 2. Interference Let S1, S2 be two sources that emit spherical sound waves in phase. S1  P d1 d2 At point P: r s1 (r ,t ) = s1 max cos(kd1 − ωt ) r s2 (r ,t ) = s2 max cos(kd2 − ωt ) ( Phase difference = k d2 − d1 S2  Destructive interference ) This is what matters… k ( d2 − d1 ) = noddπ d2 − d1 = nodd If both waves have the same amplitude (equal distance to sources), these points can even have zero intensity! λ 2
3. 3. + amplitude - amplitude
4. 4. Interference in real life? Your stereo equipment does not seem to produce these minimum intensity spots… • many frequencies at the same time • multiple reflections on walls, ceiling, furniture… DEMO: Interference
5. 5. Beats Consider two harmonic waves meeting at x = 0. Same amplitudes, but ω2 = 1.15 ω1. The displacement versus time for each is shown below: A cos(ω1t ) A cos(ω2t ) C(t) = A(t) + B(t) Constructive interference Destructive interference
6. 6. DEMO: Beats Beats (math) Video A cos(ω1t ) + A cos(ω2t ) = 2A cos ( ωLt ) cos ( ωHt ) where ωL = 1 ( ω1 − ω2 ) 2 and ωH = 1 ( ω1 + ω2 ) 2 cos(ωLt) Beat 1 Beat 3 Beat 2 Note: What you actually hear (beats) has frequency fbeat fL = = f1 − f2 2
7. 7. Doppler Shift DEMO: Whistle Even better: http://www.lon-capa.org/~mmp/applist/doppler/d.htm
8. 8. Doppler math: moving source • Speed of sound v is constant. t=0 vS (source) • Source emits λ • Listener (ear) perceives λ’ λ ' = λ − v sT t=T vST λ’ v v vs = − f' f f v f '=f v −v s λ Front of wave emitted at t = 0 Source moving with vS (vS>0 from listener to source) Stationary listener v fL = fS v +vs
9. 9. Doppler math: moving listener v (sound) t=0 v L (listener) λ vS λ ' = λ − v LT ' v v v = − L f' f f' t = T’ λ’ vLT’ v + vL f '=f v Stationary source. Source moving with vL (vL>0 from listener to source) v + vL fL = fS v
10. 10. Moving source and moving listener v +vL fL = fS v + vS vL, vS > 0 in direction from listener to source (v > 0 always) To get signs correct 1) sketch the situation, including a few wavefronts 2) decide whether observed wavelength or period will be shorter or longer 3) use this to guide whether frequency increases, decreases 4) keep in mind speed of sound does not depend on what the source or observer is doing
11. 11. ACT: Doppler A train is approaching you as you stand on a platform at a railway station. As the train approaches, it slows down. All the while, the engineer is sounding the horn at a constant frequency of 500 Hz. 1. Heard frequency is greater than 500 Hz and increases as train slows down 2. Heard frequency is greater than 500 Hz and decreases as train slows down 3. Heard frequency is less than 500 Hz and increases as train slows down 4. Heard frequency is less than 500 Hz and decreases as train slows down Source approaching listener: wavefronts are squeezed together: λ↓ Effect must be getting smaller (back to source frequency): f decreases f↑
12. 12. In-class example: Doppler A source of sound has a characteristic frequency f. The speed of sound is v. Consider the following four scenarios: 1. Static source, vobserver = v/2 toward source 2. Static source, vobserver = v/2 away from source 3. Static observer, vsource = v/2 toward observer 4. Static observer, vsource = v/2 away from observer Order f1, f2, f3, f4 from lowest to highest. A. B. C. D. E. f 1 = f2 = f3 = f4 f 2 = f4 , f1 = f3 f 1 , f2 , f3 , f4 f 2 , f4 , f1 , f3 f 4 , f3 , f2 , f1
13. 13. A source of sound has a characteristic frequency f. The speed of sound is v. Consider the following four scenarios: 1. Static source, vobserver = v/2 toward source 2. Static source, vobserver = v/2 away from source 3. Static observer, vsource = v/2 toward observer 4. Static observer, vsource = v/2 away from observer Order f1, f2, f3, f4 from lowest to highest. A. B. C. D. E. f 1 = f2 = f3 = f4 f 2 = f4 , f1 = f3 f 1 , f2 , f3 , f4 f 2 , f4 , f1 , f3 f 4 , f3 , f2 , f1 v v+ 2 = 1.5f f1 = f v f3 = f v v v− 2 = 2f f2 = f f3 = f v− v v v+ v 2 = 0.5f v 2 = 0.67f It is NOT option B: 2 and 4 (or 1 and 3) are not equivalent. You need to think about the motion relative to air, too.
14. 14. Shock waves What if the source (a plane, for instance) is moving almost at the speed of sound? http://www.lon-capa.org/~mmp/applist/doppler/d.htm Air piles up here. vsource v High pressure and density in front of plane Large aerodynamic drag (plane pushes on air, air pushes back) “Sound barrier”
15. 15. Supersonic speeds And what if vsource > v ? http://www.lon-capa.org/~mmp/applist/doppler/d.htm Points of constructive interference are along the red lines (sides of a cone) BIG amplitude there! When this cone touches the ground… vsource > v …a person on the yellow line hears a very loud sound (sonic boom)
16. 16. Mach number vsource > v vsourcet α vt sin α = vt v sourcet = v v source vs = Mach number v
17. 17. 2α ~ 130° Mach ~ 1.1