Question

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

Answer 1

the frequency of the beats is 8.7687 kHz

Explanation:

Given the data in the question;

The frequency for stationary source and moving observer is;

f' = f( 1 +( v_observer/v_sound))

we know that, speed of sound in dry air = 343 m/s

so we substitute

f' = 41.2 kHz( 1 + (33.0 m/s / 343 m/s) ) = 41.2kHz( 1 + 0.0962) = 41.2kHz(1.0962)

f' = 45.1634 kHz

Now the frequency for stationary observer and moving source with frequency f' will be

f" = f'( 1 / (1 - ( v_observer/v_sound)))

45.1634 kHz( 343 / 343 - 33)

we substitute

f" = 45.1634 kHz( 1 / (1 - (33.0 m/s / 343 m/s)))

f" = 45.1634 kHz( 1 / (1 - 0.0962))

f" = 45.1634 kHz( 1 / 0.9038 )

f" = 45.1634 kHz( 1.1064 )

f" = 49.9687 kHz

Now the beat frequency will be;

f_beat = f' - f

we substitute

f_beat = 49.9687 kHz - 41.2 kHz

f_beat = 8.7687 kHz

Therefore, the frequency of the beats is 8.7687 kHz