A source of sound with frequency $f_{0} = 1700 H z$ and an observer are at the same point. At $t = 0$ , the source starts receding from the observer with a constant acceleration $a = 10 m / s^{2}$ . The velocity of the sound in air is $V = 340 m / s$ . then chose the correct option(s).

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The sound emitted by the source at $t = 8.8 \sec$ (approximately) is received by the stationary observer at $t = 10 \sec$ .

The frequency of sound received by the stationary observer at $t = 10 \sec$ is approximately 1550 HZ

The frequency of sound received by the stationary observer at $t = 10 \sec$ is approximately 1350 HZ

The sound emitted by the source at $t = 6.8 \sec$ (approximately) is received by the stationary observer at $t = 10 \sec$ .

answer is A, D.

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Detailed Solution

Let the sound emitted by the source at time ‘t’ is received by the stationary observer at t =10 sec.
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$d = \frac{1}{2} a t^{2} = \frac{1}{2} \times 10 t^{2} = 5 t^{2}$
Now, d= 340(10-t)
$5 t^{2} = 340 (10 - t) t^{2} + 68 t - 680 = 0 t = \frac{- 68 \pm \sqrt{{(68)}^{2} + 4 \times 680}}{2}$