A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects a beat frequency of 700 Hz. The speed of the bat is closest to

3.948m/s

Explanation:

To solve this we need to apply Doppler effect theory

So

To find the frequency received by insect will be gotten when the Source and observer both are moving in same direction which is given by

f1 = f0 x (V - Vo)/(V - Vs)

f0 = 30.0 kHz

V = 344 m/s

Vs will now be the speed of the bat and

Vo will be the speed of the object which is = 0 m/s

So substituting we have

f1 = 30 x 10^3 x (344- 0)/(344- Vs)

Next to find the frequency reflected by wall we use

f2 = f1 x (V + Vs)/(V + Vo)

So substituting the value of f1 calculated above we have

f2 = 30 x 10^3 x (344 + Vs) x (344 - 0)/[(344 - Vs) x (344 + 0)]

f2 = 30 x 10^3 x (344 + Vs)/(344- Vs)

But the beat frequency detected by bat is 700 Hz,

So we say

f2 - f0 = 700 Hz

30 x 10^3 x (344+ Vs)/(344 - Vs) - 30x 10^3 = 700

(344 + Vs)/(344 - Vs) = 1 + 700/30000 = 1.023

344 + Vs = 344 x 1.023 - Vs x 1.0233

Vs = 344 x ( 1.023 - 1)/(1 + 1.023)

So finally

Vs = Speed of source that is the bat is = 3.949m/s

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