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poizon
1 month ago
8

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]1 month ago
8 0

The beat frequency is nearly 9.2 kHz

\texttt{ }

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!

\texttt{ }

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}

\texttt{ }

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}

\texttt{ }

Conclusion:

The frequency of the beats measures approximately 9.2 kHz

\texttt{ }

Learn more

  • Doppler Effect:
  • Doppler Effect Example:

\texttt{ }

Answer details

Grade: College

Subject: Physics

Chapter: Sound Waves

Ostrovityanka [3.2K]1 month ago
3 0

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/sis the velocity of the car

f=41.2 kHzis 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

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