Two sound speakers located at different points S1 and S2 are in-phase emitting a

Question

Two sound speakers located at different points S1 and S2 are in-phase emitting a sound of frequency 494 Hz travelling in air.

Six observers at points A, B, C, D, E, and F are listening to the sound created by the superposition of the two waves.

The listing below indicates the distance from each observer to speakers S1 and S2.

For example, D-S2 = 1.23 m will indicate that the observer at D is at a distance of 1.23 m from the speaker at S2.

Which of the observers will listen to no sound on account of destructive interference?

Choose which are the correct answers from the list below.


A) E-S1 = 170 cm and E-S2 = 308 cm
B) B-S1 = 210 cm and B-S2 = 279 m
C) F-S1 = 170 cm and F-S2 = 273 cm
D) C-S1 = 244 cm and C-S2 = 210 cm
E) D-S1 = 279 cm and D-S2 = 210 cm
F) A-S1 = 210 cm and A-S2 = 244 m

Explanation / Answer

For the waves to be in Destructive Interference , their phase difference should be either or odd multiples of .

Let d1 be the distance from an observer to point S1 and d2from point S2.

As we know that phase difference of the wave after travelling distance d is given by =k*d where k=2/.

Given the initial phases of two waves are same, their phase difference after reaching an observer is given by =k*(d1 - d2 ).

Since the waves are in destructive interference, we can say that

|k*(d1 - d2 )| = (2n+1)

2(d1 - d2)/ = (2n+1)

|(d1 - d2)| = (2n+1)/2

And =c/f

And based upon the velocity of sound in air, we can know which observers will listen no sound because of destructive interference.

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