The two speakers at S 1 and S 2 are adjusted so that the observer at O hears an

Question

The two speakers at S1 andS2 are adjusted so that the observer atO hears an intensity of 4 W/m2 wheneither S1 orS2 is sounded alone. They are driven inphase (at the speakers) with various frequencies of sound. Assumethat the speed of sound is 320 m/s.

a) Find the three lowest frequencies,f1 < f2 < f3, for which theobserver at O will hear an intensity of 16W/m2 when both speakers are on.

f2 = Hz   

f3 = Hz    

b) Find the three lowest frequencies,f1 < f2 < f3 , for which theobserver at O will hear no sound when bothspeakers are on.

f2 = Hz   

f3 = Hz    

c) Find the lowest frequency for whichthe observer at O will hear 8 W/m2 whenboth speakers are on.

f = Hz    

d) Find the lowest frequency for whichthe observer at O will hear 4 W/m2 whenboth speakers are on.

f = Hz

Explanation / Answer

I1 = 4 W/m2, distance OS2 = 5 m,so path difference = 5 - 4 = 1 m, v = 320 m/s = v/f the intensity at O when both speakers are on is I = 4I1cos2(/) =4I1cos2(f/v) =16cos2(f/320) a) I = 16 W/m2, so cos2(f/320) = 1 cos(f/320) = ±1 f/320 = k, where k = 0, 1, 2, ... f = 320*k f1 = 320 Hz, f2 = 640 Hz, f3 = 960Hz b) I = 0, so cos2(f/320) = 0 f/320 = (k + 0.5), where k = 0, 1, 2, ... f = 320(k + 0.5) f1 = 160 Hz, f2 = 480 Hz, f3 = 800Hz c) I = 8 W/m2, cos2(f/320) = 1/2 cos(f/320) = ±1/2 f/320 = (2k + 1)/4 f = 80(2k + 1) f1 = 80 Hz d) I = 4 W/m2, cos2(f/320) = 1/4 cos(f/320) = ±1/2 f/320 = (2k + 1)/3 f = 320(2k + 1)/3 f1 = 320/3 = 107 Hz

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