Question

A train approaching a crossing $\text{X}$ at a speed $40\text{m/s}$ sounds a horn of frequency $620\text{Hz}$, when it is at a distance of $30\text{m}$ from the crossing. Speed of sound in air is $334\text{m/s}$. What is the frequency heard by an observer (at rest) who is at a perpendicular distance of $40\text{m}$ from the crossing on the straight road.

[Assume, medium is stationary]

• A. $620\text{Hz}$
• B. $310\text{Hz}$
• C. $688\text{Hz}$
• D. $668\text{Hz}$

Answer 1

The correct option is D $668\text{Hz}$


From the figure, we deduce that,

$\cos \theta =\frac{30}{50}=\frac{3}{5}$

Given that,

Actual frequency $\left(f_0\right)=620\text{Hz}$
Velocity of source $\left(v_s\right)=40\text{m/s}$
Velocity of observer $\left(v_o\right)=0$
Velocity of sound in a stationary medium $\left(v\right)=334\text{m/s}$

Let $f_{app}$ be the apparent frequency heard by observer.

From Doppler effect, we can write that,

$f_{app}=f_0\left(\frac{v\pm v_o}{v\pm v_s}\right)$
$\Rightarrow f_{app}=f\left(\frac{v}{v-v_s\cos \theta }\right)$
[(-) ve sign is because source approaching the observer and we take the component of velocity along the line joining source and observer]

Substituting the given data, we get
$f_{app}=620\left(\frac{334}{334-40\times \frac{3}{5}}\right)=620\times \frac{334}{310}=668\text{Hz}$