A bat emits a sound at a frequency of 39.0 kHz as it approaches a wall.The bat detects beats with a frequency of 827 Hz between the sound it emits and the echo bouncing from the wall.What is the speed of the bat if the speed of sound in air is 343 m/s?
A)2.7 m/s
B)3.0 m/s
C)3.6 m/s
D)5.4 m/s
E)9.0 m/s

Question

Quizplus · Accepted Answer

The frequency of beats between the emitted sound and the echo is equal to the difference between their frequencies:

827 Hz = |f_emit - f_echo|

Since the bat is approaching the wall, the frequency of the echo will be higher than the emitted frequency.

f_emit - f_echo = -827 Hz

f_emit - (f_emit - 827 Hz) = 827 Hz

2f_emit = 1654 Hz

f_emit = 827 Hz

Now, we can use the equation for the Doppler effect to relate the emitted frequency to the speed of the bat:

f_emit = f_source / (1 + v_source/v_sound)

where v_source is the speed of the bat and v_sound is the speed of sound in air.

Substituting the given values, we get:

827 Hz = 39.0 kHz / (1 + v_source/343 m/s)

v_source = 3.6 m/s (rounded to one decimal place)

Therefore, the speed of the bat is 3.6 m/s.

A Bat Emits a Sound at a Frequency of 39