Doppler effect in multiple media

We will consider reflections at the interfaces. ( observer to source taken as positive direction ).

I have referred to $f_{0}$ as $100 Hz$ .

First let us see the $B_{1}-B_{2}$ interface Here if we assume the B_{2} interface to be an observer and the bat to be the source applying doppler effect formula we get $f = 4f_{0}/3$ .Since speed of sound is $v$ and source moves with a speed $v/4$ .

This frequency will continue all the way along till box $B_{20}$ since there is no relative motion between the blocks in between.Frequency remains unchanged only wavelength and consequently speed changes.Hence we can skip all the middle boxes and fast forward to the last box.

So now treating the $B_{19} - B_{20}$ interface as the source ( and noting that speed of sound in $B_{20}$ is $20v$ ) , we get,that the frequency observed by the bat is

$f_{1} =$ $4f_{0} /3 * 81/80 = 27 f_{0} / 20$ .

Now let us solve for the frequency from the bat inside the box $B_{n}$ noting that speed of sound inside it is $nv$ and treating the $B_{n} - B_{n+1}$ interface as observer we get frequency as $\frac{4n}{4n-1}$ $f_{0}$ .

Again we skip all the middle boxes since frequency remains unchanged all the way.Finally using the $B_{19}-B_{20}$ as the source and bat as observer,we get,frequency observed as

$f _{2}$ $=$ $\frac{4n}{4n-1}$ $f_{0}$ $* 81/80 =$ $\frac{324n}{320n-80}$ $f_{0}$ .

Knowing that beat frequency is $f_{2} - f_{1}$ = $130 Hz$

Hence solving we get,

$n = 7$ .

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Inspired by @Spandan Senapati