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A bat flies toward a wall, emitting a steady sound of frequency f in its own reference frame. The bat hears its own sound plus the sound reflected by the wall, which together produce a beat frequency of 8.00 Hz. If the bat is flying at 1.00 m/s, what is the frequency f? (You may take the speed of sound in air to be 344.0 m/s.)

Respuesta :

Answer:

1372 Hz

Explanation:

[tex]f[/tex] = actual frequency emitted by bat

Consider the wall as the listener and bat as the source of sound

[tex]v_{L}[/tex] = speed of listener = 0 m/s

[tex]v_{S}[/tex] = speed of source = [tex]v[/tex]

[tex]V[/tex] = Speed of sound = 344 m/s

[tex]f'[/tex] = frequency observed by wall

Since source is moving towards the listener, using Doppler's law

[tex]f' = \frac{(V - V_{L})f}{V - V_{s}} \\f' = \frac{(V - 0)f}{V - v}\\f' = \frac{344f}{344 - v}[/tex]

After sound gets reflected, Consider the wall as the source and bat as the listener.

[tex]v_{v}[/tex] = speed of listener(bat) = v

[tex]v_{S}[/tex] = speed of source(wall) = 0

[tex]V[/tex] = Speed of sound = 344 m/s

[tex]f''[/tex] = frequency of the reflected sound heard by bat

Since listener(bat) is moving towards the source(wall), using Doppler's law

[tex]f'' = \frac{(V + V_{L})f'}{V - V_{s}} \\f'' = \frac{(V + v)f'}{V - 0}\\f'' = \frac{(344 + v)f'}{344}\\ f'' = \frac{(344 + v)}{344}\frac{344f}{344 - v}\\ f'' = \frac{(344 + v)f}{(344 - v)}\\ f'' = \frac{(344 + 1)f}{(344 - 1)}\\ f'' = \frac{345f}{343}[/tex]

It is given that

[tex]f'' - f = 8 \\\\\frac{345f}{343} -f = 8\\\\f = 1372 Hz[/tex]

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