Question

An observer in a speeding car sees a traffic jam ahead of him. He slows down to $36\text{km/hour}$. He finds that traffic has eased. A car moving ahead of him at $18\text{km/hour}$ is honking at a frequency of $1392\text{Hz}$. If the speed of sound is $343\text{m/s}$, the frequency of the honk as heard by him will be

• A. $1332\text{Hz}$
• B. $1372\text{Hz}$
• C. $1412\text{Hz}$
• D. $1454\text{Hz}$

Answer 1

The correct option is C $1412\text{Hz}$

In this case, velocity of source $V_s=18\text{kmph}=18\times \frac{5}{18}=5\text{m/s}$
and similarly,
Velocity of observer $V_o=36\text{kmph}=10\text{m/s}$



Let $N_0$ be the frequency of sound heard by the observer, if both the source of sound and observer are at rest.
Let $v$ be the velocity of sound in the stationary medium, $v_s$ be the velocity of the source of sound and $v_o$ be the velocity of the observer.

When the source of sound and observer are moving in the same direction, frequency heard by the moving observer is given by

$N^{\prime }=N_0\left(\frac{v+v_o}{v+v_s}\right)$

Given,

Velocity of observer $v_o=36\text{km/hr}\text{or}10\text{m/s}$

Velocity of source of sound $v_s=18\text{km/hr}\text{or}5\text{m/s}$

Velocity of sound in stationary medium $v=343\text{m/s}$

Substituting the given data, we get

$N^{\prime }=1392\times \left(\frac{343+10}{343+5}\right)=1412\text{Hz}$
$\Rightarrow N^{\prime }=1412\text{Hz}$ is the frequency heard by the moving observer.