Chapter 12, Problem 72P

To determine

The beat frequency.

Answer to Problem 72P

The beat frequency is $3.2\text{kHz}$.

Explanation of Solution

The velocity of the bat is $15{\text{ms}}^{-1}$ and the emitted frequency is $35\text{kHz}$.

Consider the bat to be the source and write the expression for Doppler shift

$f_1=\frac{v-v_o}{v-v_s}f$                                                             (I)

Here, $f_1$ is the observed frequency considering bat as the source, $v_s$ is the velocity of source i.e. the bat, $v_o$ is the velocity of observer, i.e. the wall, $v$ is the velocity of the sound wave and $f_1$ is source frequency i.e. the bat.

Since the velocity of bat is in the direction of propagation of the wave, $v_s$ is considered positive. The wall is at rest and hence $v_o$ is considered zero.

Substitute $0$ for $v_o$  and $v_b$ for $v_s$ in (I). Here, $v_b$ is the velocity of the bat

$f_1=\frac{v}{v-v_s}f$                                                                   (II)

Rearrange

$f_1=\frac{1}{1-\frac{v_s}{v}}f$                                                                   (III)

Consider the bat to be the receiver and write the expression for Doppler shift

$f_2=\frac{v-{v^{\prime }}_o}{v-{v^{\prime }}_s}{f}_1$                                                            (IV)

Here, $f_2$ is the observed frequency considering bat as the receiver, ${v^{\prime }}_s$ is the velocity of source i.e. the wall, ${v^{\prime }}_o$ is the velocity of observer i.e. the bat, $v$ is the velocity of the sound wave and $f_1$ is source frequency i.e. the reflected frequency from the wall.

Since the velocity of bat is in the direction of propagation of the wave, ${v^{\prime }}_o$ is considered negative i.e. $-v_b$ and ${v^{\prime }}_s$ is considered zero because the wall is at rest.

Substitute $0$ for ${v^{\prime }}_s$  and $-v_o$ for ${v^{\prime }}_o$ in (IV)

$f_2=\frac{v-\left(-v_b\right)}{v}{f}_1$

Substitute (II) in (IV)

$f_2=\left(\frac{v+v_b}{v}\right)\frac{v}{v-v_b}f$

Rearrange

$f_2=\frac{v+v_b}{v-v_b}f$                                                  (V)

Write the expression for beat frequency between the emitted frequency and the echo.

$f_b=f_2-f$                                                (VI)

Here, $f_b$ is the beat frequency.

Substitute (VI) in (V)

$\begin{array}{c}{f}_b=\frac{v+v_b}{v-v_b}f-f\\ =\left(\frac{v+v_b}{v-v_b}-1\right)f\\ =\left(\frac{v+v_b-v+v_b}{v-v_b}\right)f\\ =\left(\frac{2v_b}{v-v_b}\right)f\end{array}$                                                       (VII)

Substitute $15{\text{ms}}^{-1}$ for $v_b$, $343{\text{ms}}^{-1}$ for $v$  and $35\text{kHz}$ for $f$ in (VII) to find $f_b$

$\begin{array}{c}{f}_2=\left(\frac{2\left(15{\text{ms}}^{-1}\right)}{343{\text{ms}}^{-1}-15{\text{ms}}^{-1}}\right)35\text{kHz}\\ =\text{3}\text{.2}\text{kHz}\end{array}$

Thus, the beat frequency is $3.2\text{kHz}$.