In the figure, two loudspeakers, separated by a distance of d 1 = 2.32 m, are in

Question

In the figure, two loudspeakers, separated by a distance of d1 = 2.32 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 d2 = 3.97 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.)

Explanation / Answer

Given that

d1 =2.32m

d2 =3.97m

The ditance between the listener and the top speaker is

x =Sqrt(d12+d22)=Sqrt(21.1433) =4.598m

Now the path difference between two sounds coming from two speakers is =x -d2 =4.598 -3.97 =0.628m

Then if the path difference is equal to lamda/2 we will get destructuve interference

then lamda/2 =0.628m then lamda =1.256m

We know that v =f*lamda

Then frequency(f) =speed/1.256m =343m/s/1.256m =273.089Hz

b)

If the path difference is equal to the wavelength then it is constuctive interference then lamda =0.628m

Then frequency =speed/0.628m =343m/s/0.628m =546.178Hz

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