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JEE Main 20207 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Evening Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2020 · 7 January · Evening Shift · Physics · Waves and Sound

A stationary observer receives sound from two identical tuning...

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2beats/sec2beats / sec . The oscillation frequency of each tuning fork is v0=1400Hz{v}_{0} = 1400Hz and the velocity of sound in air is 350m/s350m / s . The speed of each tuning fork is close to:A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes

Options

  1. A12m/s\dfrac{1}{2}m / s
  2. B1m/s1m / s
  3. C14m/s\dfrac{1}{4}m / s
  4. D18m/s\dfrac{1}{8}m / s

Correct answer

C. 14m/s\dfrac{1}{4}m / s

Step-by-step solution

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes
f0(CC−V)−f0(CC+V)=2{f}_{0}\left(\dfrac{C}{C - V}\right) - {f}_{0}\left(\dfrac{C}{C + V}\right) = 2
V=14m/sV = \dfrac{1}{4}m / s
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