Honors Classical Physics I PHY141 Lecture 32 Sound Waves Please set your clicker to channel 21 11/10/2014 Lecture 32 1 Real Loudspeakers • Bosch 36W column loudspeaker polar pattern • Monsoon Flat Panel speaker: (5 dB grid) – 400 Hz: – 1 kHz: – 3 kHz: – 7 kHz: 11/10/2014 Lecture 32 2 Example Your ears are sensitive to differences in pitch, but they are not very sensitive to differences in intensity. You are not capable of detecting a difference in sound intensity level of less than 1 dB. • By what factor does the sound intensity increase if the sound intensity level β increases from 60 dB to 61 dB ? I2 I1 2 1 (10 dB) log10 (10 dB) log10 1 dB I0 I0 I2 I 2 I1 (10 dB) log10 (10 dB) log10 1 dB I0 I0 I1 I2 100.1 ... I1 11/10/2014 Lecture 32 3 Example A 3 db Increase in sound level corresponds to: +3 dB = 10 dB log(Inew/Iold) Inew/Iold = 103/10 = 100.3 = 2.0 11/10/2014 Lecture 32 4 A 10 dB increase in the sound level corresponds to an increase in intensity of factor of … Rank Responses 1 2 3 4 5 6 Other β(I) ≡ (10 dB) log10(I /I0) with I0≡10–12 W/m2 Δβ(I) ≡ β(If) – β(Ii) = (10 dB) log10(If /Ii) 11/10/2014 Lecture 32 60 5 Standing Sound Waves • In open or closed air pipes sound waves will form standing waves when the air is rhythmically excited at the appropriate frequency… This is the basis for all wind-instruments: organs, flutes, etc. • Note that the velocity of sound in air is fixed (v = 344 m/s for normal conditions), thus the product of fλ = v is fixed, and f and λ are NOT independent! • As for standing waves on a string under tension, we expect the wavelength λ of the standing wave, and the length L of the pipe, to be similarly related. – At an OPEN END of the pipe, the pressure is closely equal to the ambient atmospheric pressure: i.e. there must be a PRESSURE NODE – NO such situation occurs for a CLOSED END; in fact there the PRESSURE has an ANTI-NODE (maximum amplitude) 11/10/2014 Lecture 32 6 Wind Pipes Considering the above statements, we arrive at the following picture (where we depict an OPEN-ENDED pipe, and graph the standing PRESSURE waves Δp(x,t) on top: L λ2/2 λ1/2 λ4/2 – A harmonic series appears (we show the 1st, 2nd, and 4th harmonic), governed by the statement that a “whole number of half-wavelengths must fit in length L of the pipe”: • i.e.: L = n(λn/2), with: n=1,2,3,… or: λn = 2L/n = λ1/n or: fn = v/λn = n v/(2L) = n f1 the DISPLACEMENT graph looks different: the place where pressure has a NODE, displacement has an ANTI-NODE and vice versa, because there the motions of the molecules (displacing themselves to keep local pressure constant) is most violent… 11/10/2014 Lecture 32 7 Stopped Wind Pipes Consider now a STOPPED pipe: a pipe with one OPEN end, and one CLOSED END (for example the clarinet), and graph the standing PRESSURE waves Δp(x,t) on top: L λ1/4 λ2/4 λ4/4 Careful inspection shows that a different harmonic series appears (we show the 1st, 3rd, and 7th harmonic): • i.e.: L = (2n – 1) λn/4 , with: n=1,2,3,… (Note: (2n – 1) is odd!) • equivalently: L = nodd λn/4 , with: n=1,3,5,… (ODD harmonics only) • or: λn = 4L/nodd = λ1/nodd or: fn = v/λn = nodd v/(4L) = nodd f1 11/10/2014 Lecture 32 8 A one-end-open, one-end-closed organ pipe is 0.34 m long; the ground tone has a frequency of … (Hz) Rank 1 2 3 4 5 6 Responses vsound = 344 m/s L Other λ1/4 11/10/2014 Lecture 32 λ2/4 60 λ4/4 9 Resonance Large displacement/pressure waves will occur when the exciting force is acting in sync with the “NATURAL FREQUENCY” of the pipe (or other type of instrument). Absent damping, the displacements can become uncomfortably large! – Resonance is used in many instruments to enhance particular (over)tones (e.g. bass reflex tube, violin/guitar body case) Example: A 0.40 m long, one-side-closed organ pipe is in exact resonance with a 0.50 m guitar string; both vibrate at the fundamental tone. – Calculate vstring: Lpipe=λ1pipe/4 and Lstring=λ1string/2 fpipe= vair/(λ1pipe=4Lpipe) = fstring= vstring/(λ1string=2Lstring) vstring=vair 2Lstring/(4Lpipe) = 0.625×344 m/s = 215 m/s – Calculate the λ and f of the sound waves in the air: λ1= 4Lpipe= 1.60 m, f1= vair/λ1=344/1.60 = 215 Hz – Calculate the λ and f of the standing wave on the guitar string: λ1=2Lstring=1.0 m; check: f1= vstring/λ1=215/1.00=215 Hz 11/10/2014 Lecture 32 10 Interference Interference is the destructive or constructive addition of displacements by waves arriving from two or more sources at a set of spatial locations. – Interference is a consequence of the SUPERPOSITION PRINCIPLE, which says that disturbances caused by individual waves at any given point simple ADD – this is a consequence of the linear character of the wave equation – Thus, a trough from one wave may coincide with an equally high peak of another, with the result there is no dis-placement at all: destructive interference. Such a “dead spot” will persist if the two waves have exactly the same frequency. – If the waves have a slightly different frequency, “beat waves” may occur: (kx 1t) (kx 2t) (kx 1t) (kx 2t) y (x,t) A cos(kx 1t) A cos(kx 2t) 2 A cos cos 2 2 1 2Acos kx 1 2 t cos 1 2 t 2Acos(kx t) cos 1 t 2Acos t cos(kx t) 2 2 2 2 – This is a traveling wave, with a frequency equal to the average frequency of the initial waves, and an amplitude which is modulated (i.e. varies in time) with a much smaller frequency equal to half the difference of the original frequencies. The INTENSITY is proportional to the amplitude squared, and thus the beat frequency we hear is simply equal to the (absolute value of) the difference in original frequencies: 1 I cos 2 t 2 11/10/2014 1 1 cos( t) double-angle formula 2 Lecture 32 11 λ maxima (twice the amplitude) Interference by SameFrequency Waves minima (dead spots) y A x B •P Interference of two synchronous equalfrequency, equalamplitude sound sources, ignoring reflections from walls, floor, ceiling, etc… Movie… Maximum in P: |AP – BP| = nλ, n=0,1,2,…; Minimum in P: |AP – BP| = (n+ ½)λ 11/10/2014 Lecture 32 12 Doppler Effect: fL=fS(v+vL)/(v+vS) Doppler effect: frequency of received sound depends on the relative motion of the source or receiver with respect to the medium (e.g. air): λ’ λ’<λ f’>f Doppler Formula: 11/10/2014 λ’’ λ v vL fL fS v vS Lecture 32 +ve for L approaching S λ’’>λ f’’<f +ve for S moving away from L 13 Doppler Effect - Derivation Doppler effect: frequency of received sound depends on the relative motion of the source or receiver with respect to the medium (e.g. air): – E.g. moving towards the source of the sound wave, my ears will “catch” more pressure variations per second than if I stay still or move away. • Thus, the frequency fL I perceive depends on my velocity vL with respect to the air. • Similarly, the motion of the sound source with respect to the air vS affects the wavelength λ of the air waves. For the simple case where all motions are along the x-direction: – Assume a sound wave in –ve x-direction; the speed of sound is v = 343 m/s λ – LISTENER who has velocity vL in +ve x-direction: vL v vL ( v ) v vL relative speed fL x – SOURCE with velocity vS in +ve x-direction: the source travels a distance vST per period, so that the effective wavelength is increased by that amount: vS v λ wavelength S vS T v vS v v S fS fS fS Combining: Doppler Formula: f L 11/10/2014 x v vL Lecture 32 v vL fS v vS +ve for L approaching S +ve for S moving away from L 14 Example v vL fL fS v vS +ve for L approaching S +ve for S moving away from L – Note, that if vL = vS then fL = fS (e.g. when the wind blows from source to receiver, nothing changes… a bat emits a high-pitched “chirp” at 80 kHz (ultra-sound) when it approached a fixed wall with velocity vBat=10 m/s – calculate the frequency of the reflected chirps the bat receives… v f reflected from wall fincident on wall f Bat v vBat v vBat v v vBat f received by Bat f reflected from wall f Bat v v vBat v v vBat 354 f Bat 80 kHz 84.8 kHz v vBat 334 – Note that the SIGN of the velocities is CRUCIAL! – the problem is even more complex when the wall is a MOVING INSECT! This is an example of SONAR 11/10/2014 Lecture 32 15 A strong wind is blowing from a stationary source towards a stationary listener … A. The received tone is lower than the emitted tone … B. The received tone is equal to the emitted tone … C. The received tone is higher than the emitted tone … 33% fL fS v vL v vS 33% +ve for L approaching S +ve for S moving away from L A. 11/10/2014 33% Lecture 32 B. C. 60 16

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