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.

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

[tex]f_B = f'-f[/tex] (1)

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

[tex]f'=\frac{v}{v-v_s}f[/tex]

where

v = 340 m/s is the speed of sound

[tex]v_s =33.0 m/s[/tex] is the speed of the car

[tex]f=41.2 kHz[/tex] is the frequency of the original sound

Substituting,

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

So, the beat frequency (1) is

[tex]f_B = 45.6 kHz - 41.2 kHz=4.4 kHz[/tex]

johanrusli johanrusli · Answer 2 · 2020-01-30T08:13:23+01:00

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:

[tex]\boxed {f' = \frac{v + v_o}{v - v_s} f}[/tex]

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!

[tex]\texttt{ }[/tex]

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:

[tex]f' = \frac{v + v_c}{v - v_c} \times f[/tex]

[tex]f' = \frac{330 + 33}{330 - 33} \times 41.2[/tex]

[tex]f' = \frac{363}{297} \times 41.2[/tex]

[tex]f' = \frac{11}{9} \times 41.2[/tex]

[tex]f' = 50 \frac{16}{45} \texttt{ kHz}[/tex]

[tex]f' \approx 50.4 \texttt{ kHz}[/tex]

[tex]\texttt{ }[/tex]

Next , we could calculate the frequency of the beats as follows:

[tex]\Delta f = f' - f[/tex]

[tex]\Delta f \approx 50.4 - 41.2[/tex]

[tex]\Delta f \approx 9.2 \texttt{ kHz}[/tex]

[tex]\texttt{ }[/tex]

Conclusion:

The frequency of the beats is about 9.2 kHz

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Learn more

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

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Answer details

Grade: College

Subject: Physics

Chapter: Sound Waves