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3 years 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.

2 answers:

iragen [17]3 years ago

8 0

The frequency of the beats is about 9.2 kHz

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall the Doppler Effect formula as follows:

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

<em>f' = observed frequency</em>

<em>f = actual frequency</em>

<em>v = speed of sound waves</em>

<em>v_o = velocity of the observer</em>

<em>v_s = velocity of the source</em>

<em>Let's tackle the problem!</em>

$\texttt{ }$

<u>Given:</u>

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

<u>Asked:</u>

frequency of the beats = Δf = ?

<u>Solution:</u>

<em>Firstly , we will calculate the observed frequency by using the formula of </em><em>Doppler Effect</em><em> as follows:</em>

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

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

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

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

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

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

$\texttt{ }$

<em>Next , we could calculate the frequency of the beats as follows:</em>

$\Delta f = f' - f$

$\Delta f \approx 50.4 - 41.2$

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

$\texttt{ }$

<h3>Conclusion:</h3>

The frequency of the beats is about 9.2 kHz

$\texttt{ }$

<h3>Learn more</h3>

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

$\texttt{ }$

<h3>Answer details</h3>

Grade: College

Subject: Physics

Chapter: Sound Waves

frosja888 [35]3 years ago

3 0

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

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

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c)

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a)

The electric force exerted on a charged particle is given by

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For a positive charge, the direction of the force is the same as the electric field.

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$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

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The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

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$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

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$F_{B_y}=qvB_y$

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And by using the right-hand rule:

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