Question

A bat, flying at 5.00 m/s, emits a chirp at 40.0 kHz. If this sound pulse is reflected by a wall, what is the frequency of the echo received by the bat

Answer 1

Answer:

The answer is below

Explanation:

Firstly, the frequency is received by the wall and then it is reflected and received by the bat.

The frequency received by the wall (f') is given by:

f' = $f(v\pm v_o)/v$

The - sign is used when the  observer is moving away from the source and + sign when the observer is moving towards the source.

Since the bat is moving towards the wall, we use a positive sign. Hence:

f' = $f(v+ v_o)/v$

The frequency reflected and received by the bat f" is:

f'' = $f'\frac{v}{ (v\pm v_s)}$

- sign is used when the source moves toward the observer and + is used when the source moves away

since the bat moves towards the wall, then::

f'' = $f'\frac{v}{(v-v_s)} =\frac{f(v+v_o)}{v}*\frac{v}{(v-v_s)} =f\frac{(v+v_o)}{(v-v_s)}$

v = speed of sound in air = 331 m/s, vo = velocity of observer = 5 m/s, vs = velocity of source = 5 m/s. Therefore:

$f"=f\frac{(v+v_o)}{(v-v_s)} =40\ kHz\frac{(331\ m/s+5\ m/s)}{(331\ m/s+5\ m/s)} \\\\f"=41\ kHz$

Answer 2

The frequency of the echo received by the bat is 38.8 kHz.

The given parameters:

• Speed of the bat, V = 5 m/s
• Actual frequency of the chirp, Fs = 40 kHz
• Speed of sound, Vs = 331 m/s

The observed frequency or frequency received by the bat is calculated by applying Doppler effect as follows;

$f_0 = f_s(\frac{v \ +/-\ v_0}{v\ +/- \ v_s} )$

Since the bat is flying away from the wall the frequency received will be smaller than the actual frequency;

$f_o = f_s (\frac{v \ - \ v_0}{v \ + \ v_s} )\\\\f_o = 40 \ kHz \times (\frac{331 - 5}{331 + 5} ) \\\\f_0 = 38.8 \ kHz$

Thus, the frequency of the echo received by the bat is 38.8 kHz.

Learn more about Doppler effect here: https://brainly.com/question/3841958