In Fig. 17-36, two loudspeakers, separated by a distance of d1 = 2.27 m, are in

Question

In Fig. 17-36, two loudspeakers, separated by a distance of d1 = 2.27 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 4.19 m directly in front of one of the speakers. Consider the audible range for normal hearing, 20 Hz to 20 kHz. What is the lowest frequency that gives the minimum signal (destructive interference) at the listener's ear? 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.)

Explanation / Answer

Path difference=sqrt(d12+d22)-d2=0.5754m

a)For destructive interference=Path difference=(2n+1)/2=0.5754

=1.1508m,0.3835m

f=v/=343/1.1508=298Hz

b)For constructive interference=Path difference=n=0.5754

=0.5754m

f=v/=596Hz

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