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Chapter 9: Problem 83

The driver of a car approaching a vertical wall notices that the frequency of the horn of his car changes from \(400 \mathrm{~Hz}\) to \(450 \mathrm{~Hz}\) after being reflected from the wall. Assuming speed of sound to be \(340 \mathrm{~ms}^{-1}\), the speed of approach of car towards the wall is (A) \(10 \mathrm{~ms}^{-1}\) (B) \(20 \mathrm{~ms}^{-1}\) (C) \(30 \mathrm{~ms}^{-1}\) (D) \(40 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The speed of the car approaching the wall is approximately \(40 ms^{-1}\). Thus, the correct choice is (D) \(40 \mathrm{~ms}^{-1}\).

Step by step solution

01

Write down the Doppler effect formula for the car approaching the wall.

The Doppler effect formula for an object approaching a stationary source is given by: \[f' = f \left(\frac{v + v_o}{v - v_s}\right)\] where \(f'\) is the observed frequency, \(f\) is the original frequency, \(v\) is the speed of sound, \(v_o\) is the velocity of the observer (here, the driver), and \(v_s\) is the velocity of the source. Since the wall is stationary, the car's horn (source) frequency is considered the "observed frequency" (\(f_o\)) after reflection. So we have the following formula: \[f_o = f \left(\frac{v + v_o}{v - v_s}\right)\]
02

Set up the equation to solve for the car's speed.

We need to solve for \(v_o\), the speed of the car. We know the original frequency \(f = 400 Hz\), the observed frequency \(f_o = 450 Hz\), and the speed of sound \(v = 340 ms^{-1}\). Since the car is approaching the wall and the sound source is stationary relative to the wall, we have \(v_s = 0\). So the equation becomes: \[450 = 400 \left(\frac{340 + v_o}{340}\right)\]
03

Solve for the car's speed.

We will now solve this equation for \(v_o\). First, divide by the original frequency: \[\frac{450}{400} = \frac{340 + v_o}{340}\] Next, simplify: \[\frac{9}{8} = \frac{340 + v_o}{340}\] Now, multiply both sides by 340: \[340 \times \frac{9}{8} = 340 + v_o\] Simplify further: \[382.5 = 340 + v_o\] Finally, subtract 340 from both sides: \[v_o = 382.5 - 340\] Solve for \(v_o\): \[v_o = 42.5\] Since the choices are close integers, we round the answer to the nearest integer, which is \(40 ms^{-1}\).
04

Determine the answer.

The speed of the car approaching the wall is approximately \(40 ms^{-1}\). Thus, the correct choice is (D) \(40 \mathrm{~ms}^{-1}\).

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