Question

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?

• A. 472 Hz
• B. 572 Hz
• C. 628 Hz
• D. 728 Hz

Answer 1

The correct option is B 572 Hz

Use equation $v_{xf}=v_{xi}+a_{x}t$ to express the speed of the source of sound:
$v_s=u_{yi}+a_{y}t=0-gt=-gt$
Now to determine the Doppler – shifted frequency heard from the falling clock radio, we have
$f^{\prime }=\left(\frac{v+v_0}{v-v_s}\right)f\Rightarrow f^{\prime }=\left[\frac{v+0}{v-(-gt)}\right]f=\left(\frac{v}{v+gt}\right)f$ . . . . (i)
$y_f=y_i+v_{yi}t-\frac{1}{2}g{t}^2\Rightarrow -15.0m=0+0-\frac{1}{2}\left(9.80m/s^2\right)t^2\Rightarrow t=1.75s$
$f^{\prime }\left[\frac{343m/s^2}{343m/s+\left(9.80m/s^2\right)\left(1.75s\right)}\right]\left(600Hz\right)=572Hz$