A source of sonic oscillations with frequency $f_{0} = 1700 H z$ and a receiver are located at the same point. At the moment $t = 0$ the source starts receding from the receiver with constant acceleration $a = 10.0 m / s^{2}$ . Assuming the velocity of sound to be equal to $v = 340 m / s$ . The oscillation frequency registered by the stationary receiver at $t = 10.0 second$ after the start of motion is 50nHz less than the natural frequency of the source. Find n.

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Detailed Solution

Single sent at time t by the source is received by the receiver.
Distance covered by the source in time t,
$S = \frac{1}{2} (10) t^{2} = 5 t^{2} m$
Time elaspsed by the sound pulse in reaching to detector $= \frac{5 t^{2}}{340} = \frac{t^{2}}{68} \sec$
$\Rightarrow t + \frac{t^{2}}{68} = 10 \Rightarrow t^{2} + 68 t - 680 = 0 \Rightarrow t = 8.8 \sec$
Velocity of the source at t, $v = a t = 88 m / s$
Thus, $f = 1700 (\frac{340}{340 + 88}) = 1350 H z$