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  • 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 8 b

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 8 beats/ sec. The oscillation frequency of each tuning fork is f_{0}=2000 \mathrm{hz} and the velocity of sound in air is 340 \mathrm{~m} / \mathrm{s}. The speed of each tuning fork is close to

Option: 1

\frac{17}{100}


Option: 2

\frac{17}{1000}


Option: 3

\frac{14}{100}


Option: 4

\frac{100}{17}


Answers (1)

best_answer

$$ \begin{aligned} & f_{1}-f_{2}=\text { beat } / \text { sec. } \\ & f_{0}\left(\frac{c}{c-v}\right)-f_{0}\left(\frac{c}{c+v}\right)=f_{1}-f_{2}=8 \\ & \operatorname{f_oc}\left(\frac{1}{c-v}-\frac{1}{c+v}\right)=8 \\ & f_{0} c\left(\frac{c+v-c+v}{c^{2}-v^{2}}\right)=8 \\ & \frac{f_{0} a \times 2 v}{a^{2}}=8 \\ & 2 v=\frac{8 c}{f_{0}} \\ & v=\frac{c}{f_{0}}=\frac{340}{2000}=\frac{17}{100} \end{aligned}

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Rishabh

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