Your clock radio awakens you with a steady and irritating sound of frequency 600Hz. One morning, you drop the clock radio out of you fourth – story dorm window, 15.0m from the ground. Assume the speed of sound is 343m/s. As you listen to the falling clock radio, what frequency do you hear just before you hear it striking the ground?

472 Hz

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572 Hz

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628 Hz

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728 Hz

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Solution

The correct option is B 572 Hz
Use equation $v_{x f} = v_{x i} + a_{x} t$ to express the speed of the source of sound:
$v_{s} = u_{y i} + a_{y} t = 0 - g t = - g t$
Now to determine the Doppler – shifted frequency heard from the falling clock radio, we have
$f^{'} = (\frac{v + v_{0}}{v - v_{s}}) f \Rightarrow f^{'} = [\frac{v + 0}{v - (- g t)}] f = (\frac{v}{v + g t}) f$ . . . . (i)
$y_{f} = y_{i} + v_{y i} t - \frac{1}{2} g t^{2} \Rightarrow - 15.0 m = 0 + 0 - \frac{1}{2} (9.80 m / s^{2}) t^{2} \Rightarrow t = 1.75 s$
$f^{'} [\frac{343 m / s^{2}}{343 m / s + (9.80 m / s^{2}) (1.75 s)}] (600 H z) = 572 H z$

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Q. Your clock radio awakens you with a steady and irritating sound of frequency 600Hz. One morning, you drop the clock radio out of you fourth – story dorm window, 15.0m from the ground. Assume the speed of sound is 343m/s. As you listen to the falling clock radio, what frequency do you hear just before you hear it striking the ground?

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