Question

Three identical sources of sound are placed on a straight line as shown in the figure . A detector D is placed at a distance d from S1. The wavelength of sound is λ.
Determine the value of d so that the intensity of the wave at the detector may be
(a) zero
(b) nine times the intensity of each source.

Answer 1

Constructive interference and destructive interference.  Intensity of the wave.

We are given identical sources. There is no phase difference. Frequency f, ω, time period T, wavelength λ and speed v are all identical.  Also their amplitudes are equal.

      k = 2π/λ = ω/v.     ω = 2πf = 2π/T = k v

A sound wave is a pressure wave and is a progressive wave. The excess pressure (compared to the equilibrium pressure) at a point is specified in the (longitudinal) wave equation as :

Sound from S1 at :
      p1 = p0 * Sin(k x - ωt)  = p0 * Sin (3d k - ωt),    as x = 3d

Sound from S2:
     p2 = p0 * Sin (kx- ωt)  = p0 * Sin (2d k - ωt),  as x = 2d

Sound from S3 :      p 3 = p0 * Sin (d k - ωt)      as x = d

Interference of the three waves at point D means the waves are added algebraically.

 Resultant wave = p1+ p2 + p3
    = 2 p0 Sin(2 k d - ωt) Cos (dk)  + p0 Sin (2d k - ωt)
    = p0 * SIn(2 d k - ωt) * [2 Cos(dk) + 1 ]

The resultant wave is zero with destructive interference, if
     Cos (dk) = -1/2
     d k = (2n+1) π  + π/3

The constructive interference occurs when
     Cos(dk) = 1 = max value.
     d k = 2 n π
     d * 2π/λ  = 2 n π
  =>  d = n λ     So integral multiple of wave length.

 Then the resultant wave is :   3 p0 Sin(2 kd - ωt)

  Since the amplitude is 3 p0 , the intensity I is (square of amplitude) 9 times the original amplitude of each source of sound.

Answer 2

Answer:

hope this answer useful to you