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Question

A small musical toy produces a steady tone at 1000 Hz1000 \mathrm{~Hz}. It is accidentally dropped from the window of a tall apartment building. How far has it fallen if the frequency heard from the window is 950 Hz950 \mathrm{~Hz} ? (Ignore air friction, and use 340 m/s340 \mathrm{~m} / \mathrm{s} for the speed of sound.)

Solution

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Answered 2 years ago
Answered 2 years ago
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From equation 12.19, i.e. the master doppler effect equation,

fL=v+vLv+vSfS(1)f_L = \frac{v+ v_L}{v+v_S} f_S \hspace{0.5 cm} (1)

Where fLf_L is the frequency of the sound heard by the listener, fSf_S is the frequency of sound produced by the source, vv is the velocity of sound, vLv_L and vSv_S are the velocities of the listener and the source respectively.

Equation 2.14 gives us the relationship between an object's height and velocity as,

v2=v022g(yy0)(2)v^2 = v_0 ^2 - 2g(y-y_0) \hspace{0.5 cm} (2)

Where v0v_0 is the intial velocity, yy and y0y_0 are the final and initial distances from the origin, and gg is the acceleration due to gravity.

Given: fS=1000f_S = 1000 Hz, fL=950f_L = 950 Hz, v=340v = 340 m/s, vL=0v_L = 0, v0=0v_0 = 0, y=0y = 0

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