Question
Two tuning forks, with natural frequency 700 Hz each, move relative to a stationary observer. Fork 1 moves towards the observer while the fork 2 moves away from the observer. Both the forks move with same velocity v on the same line. The observer, standing between the two forks, hears 4 beats per sec. Using the speed of sound in air as vs = 350 m s−1, the speed of each tuning fork is :
Options
Solution
[Key Concept: The beat frequency is the difference between the Doppler-raised frequency of the approaching source and the lowered frequency of the receding one.]
$f_{1} =$ frequency heared by observer due to tunning fork 1
$f =$ frequency heared by observer due to tunning fork 2
$$\begin{matrix} & f_{1} = f_{0}\left\lbrack \frac{v_{\omega} \pm v_{0}}{v_{\omega} \pm v_{s}} \right\rbrack \\ & \ \Rightarrow f_{1} = 700\left\lbrack \frac{350 + 0}{350 - v} \right\rbrack \\ & f_{2} = 700\left\lbrack \frac{350}{350 + v} \right\rbrack \end{matrix}$$
According to given condition,
$$\begin{matrix} & f_{\text{beat~}} = f_{1} + f_{2} \\ & 4 = 700\left\lbrack \frac{350}{350 - v} \right\rbrack - 700\left\lbrack \frac{350}{350 + v} \right\rbrack \\ & 4 = 700 \times 350\left\lbrack \frac{(350 + v) - (350 - v)}{(350)^{2} - v^{2}} \right\rbrack \\ & \ (350)^{2} - v^{2} = \frac{1}{2} \times 350 \times 350 \times 2v \\ & \ (350)^{2} - v^{2} = (350)^{2}v \\ & v^{2} + (350)^{2}v - (350)^{2} = 0 \\ & v^{2} + 122500v - 122500 = 0 \end{matrix}$$
From the given option the most appropriate choice for speed of tunning fork will be
$$v \approx 1\text{ }m/s$$
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