Question

A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects beats such that the frequency of the echo is 900 Hz higher than the frequency the bat is emitting. The speed of the bat is closest to

Answer 1

Answer:

The speed of the bat is 5.02 m/s.

Explanation:

Given that,

Frequency = 30.0 kHz

Frequency of echo = 900 Hz

We need to calculate the frequency

Using formula of beat frequency

$f_{b}=f_{1}-f_{2}$

Put the value into the formula

$900=f_{1}-30\times10^{3}$

$f_{1}=900+30000$

$f_{1}=30900\ Hz$

We need to calculate the speed of the bat

Using Doppler equation

$f_{apr}=f\times(\dfrac{v_{sound}+v_{observer}}{v_{s}-v_{source}})$

Put the value into the formula

$30900=30000\times(\dfrac{340+v_{bat}}{340-v_{bat}})$

$\dfrac{30900}{30000}=\dfrac{340+v_{bat}}{340-v_{bat}}$

$1.03=\dfrac{340+v_{bat}}{340-v_{bat}}$

$340\times1.03-340=v_{b}+1.03v_{b}$

$10.2=2.03v_{bat}$

$v_{bat}=\dfrac{10.2}{2.03}$

$v_{bat}=5.02\ m/s$

Hence, The speed of the bat is 5.02 m/s.