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.
A) 9.2 kHz
B) 4.1 kHz
C) 4.6 kHz
D) 1.2 kHz
E) 8.2 kHz

Question

Quizplus · Accepted Answer

The beat frequency is equal to the difference between the frequency of the emitted sound and the frequency of the reflected sound. The frequency of the emitted sound is 41.2 kHz. The frequency of the reflected sound is Doppler-shifted due to the motion of the car. The Doppler shift is given by the equation:$$f_d = f_s * (v + v_s) / (v + v_d)$$where:* $$f_d$$ is the Doppler-shifted frequency* $$f_s$$ is the source frequency* $$v$$ is the speed of sound* $$v_s$$ is the speed of the source* $$v_d$$ is the speed of the detectorIn this case, the source is the ultrasonic device and the detector is the policeman. The speed of the source is 0 m/s because the ultrasonic device is stationary. The speed of the detector is also 0 m/s because the policeman is stationary. Therefore, the Doppler shift is given by:$$f_d = f_s * v / v$$$$f_d = 41.2 kHz * 330 m/s / 330 m/s$$$$f_d = 41.2 kHz$$The beat frequency is equal to the difference between the frequency of the emitted sound and the frequency of the reflected sound. Therefore, the beat frequency is:$$f_b = f_s - f_d$$$$f_b = 41.2 kHz - 41.2 kHz$$$$f_b = 0 Hz$$Therefore, the frequency of the beats is 0 Hz.

A Policeman in a Stationary Car Measures the Speed of Approaching