The two sources of sound, S_{1} and S_{2}, emitting waves of equal wavelength 20.0 cm, are placed with a separation of 20.0 cm between them. A detector can be moved on a line parallel to S_{1} S_{2} and at a distance of 20.0 cm from it. Initially, the detector is equidistant from the two sources. Assuming that the waves emitted by the sources are in detector should be shifted to detect a minimum of sound.

#### Solution

Given:

Wavelength of sound wave *λ* = 20 cm

Separation between the two sources AC = 20cm

Distance of detector from source BD = 20 cm

If the detector is moved through a distance *x, *then the path difference of the sound waves from sources A and C reaching B is given by:

Path difference = AB\[-\]BC

=\[\sqrt{\left( 20 \right)^2 + \left( 10 + x \right)^2} - \sqrt{\left( 20 \right)^2 + \left( 10 - x \right)^2}\]

To hear the minimum, this path difference should be equal to :

\[\frac{\left( 2n + 1 \right)\lambda}{2}\]=\[\frac{\lambda}{2}\]= 10 cm

So,

\[\sqrt{\left( 20 \right)^2 + \left( 10 + x \right)^2} - \sqrt{\left( 20 \right)^2 + \left( 10 - x \right)^2}\]= 10

On solving, we get, *x* = 12.6 cm.

Hence, the detector should be shifted by a distance of 12.6 cm.