Q 11-14-075JEE MainJEE Main 2022 (26 Jul, Shift 1)Medium
When a car is approaching the observer, the frequency of its horn is $100\ \text{Hz}$. After passing the observer, it is $50\ \text{Hz}$. If the observer moves with the car, the frequency will be $\dfrac x3\ \text{Hz}$ where $x$ = ______.
Numerical value type. Enter your answer.
Answer: 200
$100 = f_0\dfrac{v}{v - v_s}$ and $50 = f_0\dfrac{v}{v + v_s}$. Dividing: $\dfrac{v + v_s}{v - v_s} = 2\Rightarrow v_s = \dfrac v3$.
$$f_0 = 100\times\frac{v - v/3}{v} = \frac{200}{3}\ \text{Hz}$$
Moving with the car there is no Doppler shift, so the observer hears $f_0 = \dfrac{200}{3}\ \text{Hz}$ and $x = 200$.
Solution by Sreeraj P, M.Sc Physics