Question:

Two stationary sources P and Q produce sounds of equal frequency of \(170\,Hz\). The velocity with which an observer has to move from source P towards source Q such that 8 beats are heard per second by the observer is:

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For two equal-frequency sources and a moving observer: \[ f_b=\frac{2fu}{v}. \] This shortcut is very useful in Doppler beat problems.
Updated On: Jun 18, 2026
  • \(20\,ms^{-1}\)
  • \(16\,ms^{-1}\)
  • \(4\,ms^{-1}\)
  • \(8\,ms^{-1}\)
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The Correct Option is D

Solution and Explanation

Concept: For a moving observer: \[ f'=f\left(\frac{v\pm u}{v}\right). \] Moving towards one source increases frequency and moving away from the other decreases frequency. Beat frequency is \[ f_b=|f_1-f_2|. \]

Step 1:
Observed frequencies.
Towards source Q: \[ f_Q=f\left(\frac{v+u}{v}\right). \] Away from source P: \[ f_P=f\left(\frac{v-u}{v}\right). \]

Step 2:
Calculate beat frequency.
\[ f_b = f_Q-f_P. \] \[ = f\left(\frac{v+u}{v}\right) - f\left(\frac{v-u}{v}\right). \] \[ = \frac{2fu}{v}. \]

Step 3:
Use given values.
\[ 8 = \frac{2(170)u}{340}. \] \[ 8=u. \] \[ u=8\,ms^{-1}. \] Therefore \[ \boxed{8\,ms^{-1}}. \]
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