JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    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 2 beats/sec. The oscillation frequency of each tuning fork is \[{{v}_{0}}=1400\] Hz and the velocity of sound in air 350 m/s. The speed of each tuning fork is close to:            [JEE MAIN Held on 07-01-2020 Evening]

    A) \[\frac{1}{4}\,m/s\]

    B)   \[\frac{1}{2}\,m/s\]

    C) \[1\,m/s\]         

    D) \[\frac{1}{8}\,m/s\]

    Correct Answer: A

    Solution :

    [a] \[{{f}_{1}}={{f}_{0}}\frac{c}{c-v}\]               \[{{f}_{2}}={{f}_{0}}\frac{c}{c+v}\] \[\Rightarrow \,\,2={{f}_{1}}-{{f}_{2}}={{f}_{0}}c\,\left[ \frac{1}{c-v}-\frac{1}{c+v} \right]\] \[=\,\,{{f}_{0}}c\,\frac{2v}{{{c}^{2}}\left[ 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right]}\] \[\Rightarrow \,\,\,v=\frac{2c}{2{{f}_{0}}}=\frac{350}{1400}=\frac{1}{4}\,m/s\]

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