JEE Physics 2023 Question with Solution

Given: speed of the car $v_s = 15 \, \text{m/s}$, speed of sound $v = 330 \, \text{m/s}$, change in frequency $f - f_0 = 40 \, \text{Hz}$.

Find: the frequency of the horn $f_0$.

For reflection from a stationary wall, the reflected sound heard by the driver undergoes Doppler shift twice. Using the relation shown in the solution:

Substitute the given values:

Now use the given change in frequency:

So,

Simplify:

Hence,

Therefore, the frequency of the horn is $420 \, \text{Hz}$.

Given: the source and the driver move toward the wall with speed $15 \, \text{m/s}$, the wall is stationary, and the observed increase after reflection is $40 \, \text{Hz}$.

Find: the original horn frequency $f_0$.

First, the wall acts as a stationary observer for the moving source. The frequency received at the wall is:

Then the reflected sound acts as if it comes from a stationary source at the wall, while the driver approaches it with speed $v_s$. So the frequency heard by the driver is:

Combining both steps:

The change in frequency is:

Therefore,

Now simplify the bracket:

Substitute $v = 330 \, \text{m/s}$ and $v_s = 15 \, \text{m/s}$:

So,

Thus, the horn frequency is $420 \, \text{Hz}$.