A stationary observer receives a sound of frequency $f_0=2000 Hz$. The apparent frequency $f$ varies with time as shown in figure. Speed of sound $=300 m/s$. Choose the correct alternative (s)?

The graph shows the situation shown in figure. The observed frequency will initially be more than the natural frequency. When the source is $P$, observed frequency is equal to its natural frequency i.e., 2000 Hz.

$$\begin{gathered} \mathrm{For} \mathrm{region} \mathrm{AP}: f=f_0\left(\frac{v}{v-v_s\cos \theta }\right) \\ \mathrm{For} \mathrm{PB}: f=f_0\left(\frac{v}{v+v_s\cos \theta }\right) \end{gathered}$$

Minimum value of $f$ will be. 

$$\begin{gathered} f_{min}=f_0\left(\frac{v}{v+v_s}\right) \mathrm{when} \cos \theta =1 \\ \mathrm{or} 1800=2000\left(\frac{300}{300+v_s}\right) \end{gathered}$$

Solving this, we get, $v_s=$33.33 m/s and maximum value of $f$ can be

$$\begin{gathered} f_{max}=f_0\left(\frac{v}{v-v_s}\right) \mathrm{when} \cos \theta =1 \\ \mathrm{or} f_{max}=2000\left(\frac{300}{300-33.33}\right)=2250 \mathrm{Hz} \end{gathered}$$