A bat flies at a speed of 6 m/s between two parallel walls while...

The bat acts as both a moving source and a moving observer.

For the wall the bat is approaching, the frequency of the reflected wave heard by the bat is given by the Doppler effect formula:

f_1 = f \left(\dfrac{v + v_b}{v - v_b} \right)

Substituting $v = 330 \text{ m/s}$, $v_b = 6 \text{ m/s}$, and $f = 4.5 \times 10^4 \text{ Hz}$:

f_1 = 4.5 \times 10^4 \left(\dfrac{330 + 6}{330 - 6} \right) =4.5 \times 10^4 \left(\dfrac{336}{324} \right) =4.5 \times 10^4 \left(\dfrac{28}{27} \right) \approx$4.67 \times 10^4 \text{ Hz}$

For the wall the bat is moving away from, the frequency of the reflected wave heard by the bat is:

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

$f_2 = 4.5 \times 10^4 \left( \dfrac{330 - 6}{330 + 6} \right) = 4.5 \times 10^4 \left( \dfrac{324}{336} \right) = 4.5 \times 10^4 \left( \dfrac{27}{28} \right) \approx 4.34 \times 10^4 \text{ Hz}$

The beat frequency is the difference between the two frequencies heard by the bat:

$f_{\text{beat}} = f_1 - f_2 = 4.5 \times 10^4$\left( \dfrac{28}{27} - \dfrac{27}{28} \right)<br><br>f_{\text{beat}} = 4.5 \times 10^4 \left( \dfrac{784 - 729}{756} \right) =45000 \times $\dfrac{55}{756} \approx$3274 \text{ Hz} \approx 3270 \text{ Hz}