91Ó°ÊÓ

The beat frequency of the sound is given as ${\mathbf{f}}_{\mathbf{beat}}\mathbf{=}{\mathbf{f}}_{\mathbf{b}}\mathbf{L}\mathbf{-}{\mathbf{f}}_{\mathbf{L}}$were, ${\mathbf{f}}_{\mathbf{beat}}$is the beat frequency of the sound, ${\mathit{f}}_{\mathbf{b}\mathbf{L}}$is the frequency of bat as a listener, ${\mathit{f}}_{\mathbf{L}}$ is the frequency of the wall as a listener.

The frequency of the wall as a listener is ${\mathit{f}}_{\mathbf{L}}\mathbf{=}\mathit{v}\mathbf{-}{\mathit{v}}_{\mathbf{b}}{\mathit{f}}_{\mathbf{b}}\mathit{s}$were v is the speed of the sound in air. ${\mathit{f}}_{\mathbf{b}}$is the speed of the bat, ${\mathit{f}}_{\mathbf{bs}}$is the frequency of the bat as a source

Calculate the frequency of the bat as listener as${\mathrm{f}}_{\mathrm{B}}\mathrm{L}=\mathrm{v}+{\mathrm{v}}_{\mathrm{b}}{\mathrm{vf}}_{\mathrm{b}}\mathrm{s}$

Substitute role="math" localid="1664347516114" $\mathrm{v}-{\mathrm{v}}_{\mathrm{b}}{\mathrm{vf}}_{\mathrm{bs}}$for $f_L$and for role="math" localid="1664347666261" $\mathrm{v}+{\mathrm{v}}_{\mathrm{b}}{\mathrm{vf}}_{\mathrm{bs}}$in equation role="math" localid="1664347557599" $f_{beat}=f_{b}L-f_L$to find $f_{beat}$

${\mathrm{f}}_{\mathrm{beat}}=\mathrm{v}+{\mathrm{v}}_{\mathrm{b}}{\mathrm{vf}}_{\mathrm{bs}}-\mathrm{v}-{\mathrm{v}}_{\mathrm{bv}}{\mathrm{f}}_{\mathrm{bs}}$

$$\begin{gathered} =\left(2\mathrm{vbv}-\mathrm{vb}\right)\mathrm{FLvb} \\ =\mathrm{v}{\mathrm{f}}_{\mathrm{beat}}2\mathrm{fL}+{\mathrm{f}}_{\mathrm{beat}} \end{gathered}$$

Substitute Substitute 344 m/s for v , 8.0 Hz for f beat and 1.70 kHz for fL in the equation (Il) to find vb.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{b}}=\left(344\mathrm{m}/\mathrm{s}\right)\left(8.0\mathrm{Hz}\right)2\left(1.70\mathrm{kHz}\right)\left(1000\mathrm{Hz}\right)+8.0\mathrm{Hz} \\ =0.808\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the bat in order hears a beat frequency of 8.0 Hz is 0.808 m/s