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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Physics
Chapter
Waves and Sound

BITSAT 2019 · Physics · Waves and Sound

Two sources of sound A and B are moving towards and away from a...

Two sources of sound AA and BB are moving towards and away from a stationary observer with the same speed respectively. If the frequency of the sound produced by both the sources is equal to 400400 Hz, then the speed of sources (approximately) if the observer detects 4 beats per second, is [Given, speed of sound =340ms−1= 340ms^{- 1}]

Options

  1. A1.7ms−11. 7ms^{- 1}
  2. B3.4ms−13. 4ms^{- 1}
  3. C2.4ms−12. 4ms^{- 1}
  4. D1ms−11ms^{- 1}

Correct answer

A. 1.7ms−11. 7ms^{- 1}

Step-by-step solution

Given, frequency of sound produced by sources AA~and BB,~fA=fB=400Hz{f}_{A} = {f}_{B} = 400Hz

Speed of sound in air, v=340ms−1v = 340ms^{- 1}

Apparent frequency heard by the observer when source AA is moving towards him,

fA′=vv−vS×fA=340340−vS×400{f}_A^{'} = \dfrac{v}{v - {v}_{S}} \times {f}_{A} = \dfrac{340}{340 - {v}_{S}} \times 400

Apparent frequency heard by the observer when source BB is moving away from him,

fB′=vv+vs×fB=340340+vs×400{f}_B^{'} = \dfrac{v}{v + {v}_{s}} \times {f}_{B} = \dfrac{340}{340 + {v}_{s}} \times 400

Given, Beats~=fA′−fB′=4= {f}_A^{'} - {f}_B^{'} = 4

340340−vs×400−340340+vs×400=4\dfrac{340}{340 - {v}_{s}} \times 400 - \dfrac{340}{340 + {v}_{s}} \times 400 = 4

⇒34000×2vS3402−vS2=1⇒ \dfrac{34000 \times 2{v}_{S}}{{340}^{2} - {v}_S^{2}} = 1

34000×2vS3402=1\dfrac{34000 \times 2{v}_{S}}{3402} = 1~ & \left[∵ {v}_S^{2} < < {340}^{2}\right]</p><p></p><p>⇒ {v}_{s} = 1 . 7ms^{- 1}$

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