A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with

Question

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A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with

a frequency of 41.2 khz. A car is approaching him at a speed of 33.0 m/s. The wave is reflected by the car and interferes with the emitted sound producing beats. What is the frequency of the beats? The speed of sound in air is 330 m/s.

Physics

2 answers:

Yuliya22 [3.3K] · Answer 1 · 2026-03-14T06:57:18+03:00

The beat frequency is nearly 9.2 kHz

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Further explanation

Recalling the Doppler Effect formula expressed as:

$\boxed {f' = \frac{v + v_o}{v - v_s} f}$

f' = observed frequency

f = true frequency

v = velocity of sound waves

v_o = observer's speed

v_s = source's speed

Let’s proceed with the analysis!

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Here are the specifics:

true frequency = f = 41.2 kHz

car speed = v_c = 33.0 m/s

sound speed in air = v = 330 m/s

Required:

beat frequency = Δf =?

Methodology:

First, we will determine the observed frequency via the formula of Doppler Effect as follows:

$f' = \frac{v + v_c}{v - v_c} \times f$

$f' = \frac{330 + 33}{330 - 33} \times 41.2$

$f' = \frac{363}{297} \times 41.2$

$f' = \frac{11}{9} \times 41.2$

$f' = 50 \frac{16}{45} \texttt{ kHz}$

$f' \approx 50.4 \texttt{ kHz}$

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Then, we’ll calculate the frequency of the beats as shown:

$\Delta f = f' - f$

$\Delta f \approx 50.4 - 41.2$

$\Delta f \approx 9.2 \texttt{ kHz}$

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Conclusion:

The frequency of the beats measures approximately 9.2 kHz

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Learn more

Doppler Effect:
Doppler Effect Example:

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Answer details

Grade: College

Subject: Physics

Chapter: Sound Waves

Ostrovityanka [3.2K] · Answer 2 · 2026-03-14T06:55:42+03:00

Answer:

The beats frequency measures approximately

4.4 kHz

Explanation:

The beat frequency arises from the original ultrasound frequency, $f=41.2 kHz$ , and the frequency of the sound reflected off the car, $f'$ :

$f_B = f'-f$ (1)

To calculate the frequency of the reflected sound, we apply the Doppler effect formula:

$f'=\frac{v}{v-v_s}f$

where

v = 340 m/s, the speed of sound

$v_s =33.0 m/s$ is the velocity of the car

$f=41.2 kHz$ is the frequency of the sound emitted

By substituting values,

$f'=\frac{340 m/s}{340 m/s-33.0 m/s}(41.2 kHz)=45.6 kHz$

Thus, the beat frequency (1) is

$f_B = 45.6 kHz - 41.2 kHz=4.4 kHz$