A stationary observer receives a sound of frequency f₀ = 2000 Hz when the source is at rest. When the source starts moving with a constant velocity starting from a large distance, the apparent frequency f varies with times as shown in figure. Speed of sound =300 m/s. Choose the CORRECT alternative (s): -

 

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

For region AP :  $f=f_0\left(\frac{v}{v-v_s\cos \theta }\right)$
For PB :  $f=f_0\left(\frac{v}{v+v_s\cos \theta }\right)$
Minimum value of f will be : 
$f_{\min }=f_0\left(\frac{v}{v+v_s}\right)$  when cos $\theta =\text{ 1}$ 
Or  $1800=2000\left(\frac{3000}{300+v_s}\right)$
Solving this we get,  ${\text{v}}_{\text{s}}=\text{ 33}.\text{33 m}/\text{s}$
And maximum value of f can be
 $f_{\max }=f_0\left(\frac{v}{v-v_s}\right) when \cos \theta =\text{ 1}$  or  $f_{\max }=2000\left(\frac{300}{300-33.33}\right)=2250Hz$