Your clock radio awakens you with a steady and irritating sound of frequency 600 Hz. One morning, it malfunctions and cannot be turned off. In frustration, you drop the clock radio out of your fourth-story dorm window, 15.0 m from the ground. Assume the speed of sound is 343 m/s. (A) As you listen

Question

Your clock radio awakens you with a steady and irritating sound of frequency 600 Hz. One morning, it malfunctions and cannot be turned off. In frustration, you drop the clock radio out of your fourth-story dorm window, 15.0 m from the ground. Assume the speed of sound is 343 m/s.

(A) As you listen to the falling clock radio, what frequency do you hear just before you hear the radio striking the ground?

(B) At what rate does the frequency that you hear change with time just before you hear the radio striking the ground?

Accepted Answer

(A) In conceptualizing the problem, note that the speed of the radio increases as it falls. Thus, it is a source of sound moving away from you with an increasing speed. We categorize this problem as one in which we must combine our understanding of falling objects with that of the frequency shift due to the Doppler effect. To analyze the problem, we identify the clock radio as a moving source of sound for which the Doppler-shifted frequency is given by

[latex] f^{\prime}=\left(\frac{v}{v-v_{S}} ight) f [/latex].

The speed of the source of sound is given by Equation 2.9 for a falling object:

[latex] v_{x f}=v_{x i}+a_{x} t \quad ext { (for constant } a_{x} ext { ) } [/latex] (2.9).

[latex] v_{S}=v_{y i}+a_{y} t=0-g t=-g t [/latex].

Thus, the Doppler-shifted frequency of the falling clock radio is

(1) [latex] f^{\prime}=\left(\frac{v}{v-(-g t)} ight) f=\left(\frac{v}{v+g t} ight) f [/latex].

The time at which the radio strikes the ground is found from Equation 2.12:

[latex] x_{f}=x_{i}+v_{x i} t+\frac{1}{2} a_{x} t^{2} [/latex] [latex] ext { (for constant } a_{x} ext { ) } [/latex] (2.12).

[latex] y_{f}=y_{i}+v_{y i} t-\frac{1}{2} g t^{2} [/latex].

[latex] -15.0 m =0+0-\frac{1}{2}\left(9.80 m / s ^{2} ight) t^{2} [/latex]

t = 1.75 s.

Thus, the Doppler-shifted frequency just as the radio strikes the ground is

[latex] f^{\prime}=\left(\frac{v}{v+g t} ight) f [/latex].

[latex] =\left(\frac{343 m / s }{343 m / s +\left(9.80 m / s ^{2} ight)(1.75 s )} ight)(600 Hz ) [/latex].

= 571 Hz.

(B) The rate at which the frequency changes is found by differentiating Equation (1) with respect to t :

[latex] \frac{d f^{\prime}}{d t}=\frac{d}{d t}\left(\frac{v f}{v+g t} ight)=\frac{-v g}{(v+g t)^{2}} f [/latex].

[latex] =\frac{-(343 m / s )\left(9.80 m / s ^{2} ight)}{\left[343 m / s +\left(9.80 m / s ^{2} ight)(1.75 s ) ight]^{2}}(600 Hz ) [/latex].

=-15.5 Hz/s.

To finalize this problem, consider the following What If?

Question 17.5: Your clock radio awakens you with a steady and irritating so...

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