A bat flying at 4.54 m/s emits a chirp at 115 kHz. If the pulse is reflected by a wall, what is the frequency (kHz) of the echo heard by the bat? The temperature of the air is 25.0 C. (HINT: What is the source of the sound heard by the bat?)
To determine the frequency of the echo heard by the bat after the pulse is reflected by a wall, we need to consider the Doppler effect. The Doppler effect is the change in frequency or pitch of a sound wave caused by the motion of the source or the observer.
In this scenario, we know that the bat is flying at a constant speed of 4.54 m/s (source) and emitting a chirp at 115 kHz. The pulse is reflected by a stationary wall (observer). We need to find the frequency of the echo heard by the bat.
The Doppler effect equation for frequency is given by:
f' = f * (v + vo) / (v + vs)
Where:
f' = frequency of the echo heard by the bat
f = frequency of the emitted chirp by the bat
v = speed of sound in air
vo = velocity of the observer (wall) relative to the medium
vs = velocity of the source (bat) relative to the medium
First, let's calculate the speed of sound in air. The speed of sound in dry air at 25.0 °C is approximately 343 m/s.
Now let's substitute the given values into the equation:
f' = 115 kHz * (343 m/s + 0 m/s) / (343 m/s - 4.54 m/s)
Simplifying the equation:
f' = 115 kHz * 343 m/s / 338.46 m/s
f' ≈ 116.78 kHz
Therefore, the frequency of the echo heard by the bat would be approximately 116.78 kHz.