Question

A train approaching a railway crossing at a speed of $120{\text{kmh}}^{-1}$ sounds a short whistle at frequency $640\text{Hz}$, when it is $300\text{m}$ away from the crossing. The speed of sound in air is $340{\text{ms}}^{-1}$. What will be frequency heard by a person standing on a road perpendicular to the track through the crossing at a distance of $400\text{m}$ from the crossing ?

• A. $660\text{Hz}$
• B. $680\text{Hz}$
• C. $700\text{Hz}$
• D. $720\text{Hz}$

Answer 1

The correct option is B $680\text{Hz}$


Given that, the speed of the source,

$v_s=120{\text{kmh}}^{-1}=\frac{120\times 5}{18}=\frac{100}{3}{\text{ms}}^{-1}$

Original Frequency of the source $f_o=640\text{Hz}$

Speed of sound, $v=340{\text{ms}}^{-1}$

From the diagram it is clear, the person receives the sound of the whistle in a direction $\text{BA}$ making an angle $\theta$ with the track.

$\cos \theta =\frac{300}{\text{AB}}$

$\text{AB}=\sqrt{(300{)}^2+(400{)}^2}=500$

$\cos \theta =\frac{300}{500}=\frac{3}{5}$

The component of velocity of the source (the train) along direction $\text{AB}$ is

$v_s\cos \theta =\frac{100}{3}\times \frac{3}{5}=20{\text{ms}}^{-1}$

The source is approaching the person with this speed component,

the frequency heard by the observer is given by,

$f=f_o\left(\frac{v+v_0}{v-v_s\cos \theta }\right)$

Since observer (person) is at rest, $v_0=0$

$\Rightarrow f=640\left(\frac{340}{340-20}\right)$

$\therefore f=680\text{Hz}$

Hence, option $\left(\text{B}\right)$ is correct.