#### Answer

The beat frequency is $3201~Hz$

#### Work Step by Step

Let $v$ be the speed of sound in the air and let $v_b$ be the speed of the bat.
To find the frequency received by the wall, we can use the equation for the Doppler effect when the source is approaching:
$f_o = \left(\frac{v}{v-v_b}\right)~f$
$f_o = \left(\frac{343~m/s}{343~m/s-15~m/s}\right)~(35,000~Hz)$
$f_o = 36,600.6~Hz$
To find the reflected frequency, we can let $f_o$ be the sound source and use the equation for the Doppler effect when the observer is approaching:
$f_r = \left(\frac{v+v_b}{v}\right)~f_o$
$f_r = \left(\frac{343~m/s+15~m/s}{343~m/s}\right)~(36,600.6~Hz)$
$f_r = 38,201~Hz$
The beat frequency is the difference between the emitted frequency and the reflected frequency. We can find the frequency difference:
$38,201~Hz - 35,000~Hz = 3201~Hz$
The beat frequency is $3201~Hz$