Chapter 12, Problem 86GP

To determine

The frequency of the wave detected by the bat.

Answer to Problem 86GP

  $54.9\text{kHz}$

Explanation of Solution

Given:

The velocity of bat, ${\mathrm{V}}_{\mathrm{b}}=6.5\text{m/s}$

The velocity of moth, ${\mathrm{V}}_{\text{moth}}=5.0\text{m/s}$

The frequency of bat, ${\mathrm{f}}_{\text{bat}}=51.35\text{kHz}$

Formula used:

The apparent frequency is calculated in two cases.

In the first case, the emitted sound with the bat as a source and moth is an observer.

Case 1:

  ${\mathrm{f}}_{\text{moth}}\text{'}=\frac{\mathrm{V}+{\mathrm{V}}_{\text{moth}}}{\mathrm{V}-{\mathrm{V}}_{\text{bat}}}{\mathrm{f}}_{\text{bat}}$

In the second case, the reflected sound of moth is the source and bat is the observer.

Case 2:

  ${\mathrm{f}}_{\text{bat}}\text{'}\text{'}=\frac{\mathrm{V}+{\mathrm{V}}_{\text{bat}}}{\mathrm{V}-{\mathrm{V}}_{\text{moth}}}\mathrm{f}{\text{'}}_{\text{moth}}$

Here, ${\mathrm{V}}_{\text{bat}}$ is the velocity of bat,

  ${\mathrm{V}}_{\text{moth}}$ is the velocity of moth,

  ${\mathrm{f}}_{\text{bat}}$ is the frequency of bat.

Calculation:

Substitute the given values:

  $\begin{array}{c}{\mathrm{f}}_{\text{moth}}\text{'}=\frac{343+5}{343-6.5}(51.35)\\ =\frac{348}{336.5}(51.35)\\ =53.105\text{kHz}\\ {\mathrm{f}}_{\text{bat}}\text{'}\text{'}=\frac{343+6.5}{343-5}\left(53.105\right)\\ =\frac{349.5}{338}(53.105)\\ =54.9\text{kHz}\end{array}$

Conclusion:

The frequency of the wave detected by the bat is $54.9\text{kHz}$ .