In the figure, two loudspeakers, separated by a distance of d_1 = 2.65 m, are in

Question

In the figure, two loudspeakers, separated by a distance of d_1 = 2.65 m, are in phase. Assume the amplitudes of the sound from the speakers are approximately the same at the position of a listener, who is d_2 = 3.92 m directly in front of one of the speakers. Consider the audible range for normal hearing, 20 Hz to 20 kHz. (a) What is the lowest frequency that gives the minimum signal (destructive interference) at the listener's ear? (b) What is the lowest frequency that gives the maximum signal (constructive interference) at the listener's ear? (Take the speed of sound to be 343 m/s.) (a) Number Units. (b) Number Units.

Explanation / Answer

a) path difference in the waves travelling from speaker 1 and speaker 2, p = sqroot(d2^2 + d1^2) - d2

given, d1 = 2.65 m

d2 = 3.92 m

so, p = 0.81169 m

now speed of sound = v = 343 m/s

let frequency of sound be f

also, for destructive interference

p = (2n-1)lambda/2

so, 0.81169 = (2n-1)lambda/2

lambda = 2*p/(2n - 1)

also

v = lambda *f

so, 343 = 2*p*f/(2n - 1)

for lowest frequerncy, n has to be lowest

n = 1

f = 343/2p = 211.287 Hz

b) for maximum signal

p = nlambda

so, 0.81169 = nlambda

lambda = p/n

also

v = lambda *f

so, 343 = p*f/n

for lowest frequerncy, n has to be lowest

n = 1

f = 343/p = 422.57 Hz

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