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Karnataka Board PUCPUC Science Class 11

The Two Sources of Sound, S1 and S2, 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 S1 S2

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Question

The two sources of sound, S1 and S2, 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 S1 S2 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.

Sum
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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.

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Chapter 16: Sound Waves - Exercise [Page 354]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 16 Sound Waves
Exercise | Q 30 | Page 354

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