In Fig. 17-35, S1 and S2 are two isotropic point sources of sound. They emit waves in phase at wavelength0.50m ; they are separated byD=1.60m . If we move a sound detector along a large circle centered at the midpoint between the sources, at how many points do waves arrive at the detector (a) exactly in phase

and (b) exactly out of phase?

Short Answer

Expert verified
  1. Number of points at which waves arrive at the detector exactly in phase are 14
  2. Number of points at which waves arrive at the detector exactly out of phase are 12

Step by step solution

01

The given data

  1. Wavelength of sound emitted isotropically by sources,λ=0.50m
  2. Separation of the two isotropic points,D=1.60m
02

Understanding the concept of the interference

Find the value of constant, which is the ratio of path difference and wavelength. Then compare it with the condition for constructive and destructive interference, we can find the number of points at which waves arrive at the detector exactly in phase and out of phase.

Formulae:

  1. ϕ=m2πfor constructive interference m=0,1,2..
  2. ϕ=(2m+1)π for destructive interference m=0,1,2,3..
03

a) Calculation of points where waves are exactly in phase

For constructive interference, the position values are given as:

q=Dλ=±1,±2,.

In this case,

q=1.600.50=3.20

The sound waves would be interfering in the upper half as well as the lower half through which the detector is moving. In the upper half, we would have positive values of x-coordinates as well as negative values of x-coordinates because the center of the circle through which the detector is moving is at the center of two sources. From the condition of constructive interference, we can conclude that phase difference is integer times wavelength.

For the positiveinterference occurs when values are betweenq=0andq=3.2 .

So the values are0,1,2,3.

For negative, interference occurs when values are3,2,1.

So, for the upper half of the circle, interference occurs at total 7 points. Similarly, at the lower half of the circle, interference would occur at 7 points.

Thus, the total points at which two waves interfere constructively are 14.

04

b) Calculation of points where waves are exactly out of phase

If two waves are exactly out of phase, it means there will be destructive interference. Phase difference for such a case using formula (ii) is ϕ=(2m+1)π.

Value of m should be below q=3.2. Therefore, points for destructive interference

conditions in the upper half are 2.5,1.5,0.5,0.5,1.5,2.5.

Similarly, for lower half, such points are2.5,1.5,0.5,0.5,1.5,2.5.

Thus, the total points at which two waves interfere destructively are 12.

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