A tuning fork of frequency 380 Hz is moving towards a wall with a velocity of 4m/s. Then the number of beats heard by a stationary listener between direct and reflected sounds will be(velocity of sound in air = 340 m/s)

No Doppler effect
n¹ = n$\Rightarrow ∆$n = 0

For sound approaching the wall, the apparent frequency will be,

$\begin{array}{l}{f}^{\mathrm{\prime }}=f_0\left(\frac{v+v_g}{v-v_s}\right)\\ \Rightarrow f^{\mathrm{\prime }}=380\times \left(\frac{340+0}{340-4}\right)\\ \Rightarrow f^{\mathrm{\prime }}=384\mathrm{Hz}\dots \dots \dots .\left(1\right)\end{array}$

This will be the frequency of sound reflected by the wall. And as the listener is also at rest, so the same frequency will be heard by it after reflection.
Also, for sound approaching the stationary listener directly, the apparent frequency will be,

$\begin{array}{l}{f}^{\prime \prime }=f_0\left(\frac{v+v_g}{v-v_s}\right)\\ \Rightarrow f^{\prime \prime }=380\times \left(\frac{340+0}{340-4}\right)\\ \Rightarrow f^{\prime \prime }=384\mathrm{Hz}\dots \dots \dots ..\left(2\right)\end{array}$

Hence, zero beats per second will be heard by stationary listener produced due to direct and reflected sounds.