Question

A person driving car at a constant speed of 15 m/s is approaching a vertical wall. The person notices a change of 40 Hz in the frequency of his car’s horn upon reflection from the wall. The frequency of horn is ____ Hz.

Answer 1

Correct Answer: 420

$f= f_{0}(\frac{V+V_{c}}{V-V_{c}})$
$f= f_{0}(\frac{330+15}{330-15})$
$f= f_{0}\frac{345}{315}$
$f- f_{0}=40$
$f_{0}(\frac{345-315}{315})=40$
$f_{0}=\frac{40\times315}{30}$
$f_{0}=420Hz$

So, the answer is $420\ Hz$

Answer 2

Doppler Effect Problem for Moving Source and Stationary Observer 

Step 1: Doppler Effect Formula for Moving Source and Stationary Observer

The observed frequency $f'$ when a source emitting frequency $f_0$ is moving towards a stationary observer at speed $v_s$ is given by:

$f' = \frac{v}{v - v_s} f_0$

where: - $v$ is the speed of sound (assumed to be 330 m/s), - $v_s$ is the speed of the source.

Step 2: Doppler Effect for Reflection

In this case, the sound is reflected off a wall, which acts as a stationary "observer" first. Then, the reflected sound with frequency $f'$ acts as a source moving towards the driver (now the observer) at speed $v_s$. So, the frequency heard by the driver $f''$ is:

$f'' = \frac{v + v_o}{v} f' = \frac{v + v_s}{v - v_s} f_0$

where we’ve used $v_o = v_s$ since the driver is approaching the wall.

Step 3: Change in Frequency

The change in frequency is given as 40 Hz:

$f'' - f_0 = 40$

Substitute the expression for $f''$:

$\frac{v + v_s}{v - v_s} f_0 - f_0 = 40$

Simplify the equation:

$\frac{v + v_s - (v - v_s)}{v - v_s} f_0 = 40$

Simplify further:

$\frac{2v_s}{v - v_s} f_0 = 40$

Step 4: Calculating $f_0$

Given that $v_s = 15 \, \text{m/s}$ and $v = 330 \, \text{m/s}$, we can solve for $f_0$:

$\frac{2 \times 15}{330 - 15} f_0 = 40$

Simplify:

$\frac{30}{315} f_0 = 40$

Solve for $f_0$:

$f_0 = 40 \times \frac{315}{30} = 420 \, \text{Hz}$

Conclusion:

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