Question

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.

Answer 1

The frequency of the beats is about 9.2 kHz

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Further explanation

Let's recall the Doppler Effect formula as follows:

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

f' = observed frequency

f = actual frequency

v = speed of sound waves

v_o = velocity of the observer

v_s = velocity of the source

Let's tackle the problem!

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Given:

actual frequency = f = 41.2 kHz

velocity of the car = v_c = 33.0 m/s

speed of sound in air = v = 330 m/s

Asked:

frequency of the beats = Δf = ?

Solution:

Firstly , we will calculate the observed frequency by using 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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Next , we could calculate the frequency of the beats as follows:

$\Delta f = f' - f$

$\Delta f \approx 50.4 - 41.2$

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

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Conclusion:

The frequency of the beats is about 9.2 kHz

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Learn more

• Doppler Effect : brainly.com/question/3841958
• Example of Doppler Effect : brainly.com/question/810552

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Answer details

Grade: College

Subject: Physics

Chapter: Sound Waves

Answer 2

Answer:

4.4 kHz

Explanation:

The frequency of the beats is given by the frequency of the original ultrasound, $f=41.2 kHz$, and the frequency of the ultrasound reflected back from the car, $f'$:

$f_B = f'-f$ (1)

The frequency of the reflected wave can be found by using the Doppler effect formula:

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

where

v = 340 m/s is the speed of sound

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

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

Substituting,

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

So, the beat frequency (1) is

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