Answer:
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
