Doppler effect: 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.

Question

Quizplus · Accepted Answer

The beat frequency is equal to the difference between the observed frequency and the emitted frequency. The observed frequency is given by:$$f_{obs} = f_e \frac{v + v_o}{v}$$where:* $f_e$ is the emitted frequency* $v$ is the speed of sound* $v_o$ is the speed of the observerThe emitted frequency is 41.2 kHz, the speed of sound is 330 m/s, and the speed of the observer is 33.0 m/s. Substituting these values into the equation, we get:$$f_{obs} = 41.2 kHz \frac{330 m/s + 33.0 m/s}{330 m/s} = 45.3 kHz$$The beat frequency is equal to the difference between the observed frequency and the emitted frequency:$$f_{beat} = f_{obs} - f_e = 45.3 kHz - 41.2 kHz = 4.1 kHz$$Therefore, the correct answer is A.

Doppler Effect: a Policeman in a Stationary Car Measures the Speed