Question

A source of sound emitting a $1200\text{Hz}$ note , travels along a straight line at a speed of $170\text{m/s}$. A detector is placed at a distance of $200\text{m}$ from the line of motion of the source. Frequency of sound received by the detector at the instant when the source gets closest to it will be [velocity of sound in air $=340\text{m/s}]$

[Assume, medium is stationary]

• A. $1400\text{Hz}$
• B. $1600\text{Hz}$
• C. $1800\text{Hz}$
• D. $1200\text{Hz}$

Answer 1

The correct option is B $1600\text{Hz}$

Sound produced by the source at $\text{A}$ will reach to the observer when the source gets closest to the observer i.e at $\text{B}$.



So, time $t$ taken by source to reach from $\text{A}$ to $\text{B}$ will be equal to time taken by sound to reach from $\text{A}$ to observer.

Hence, $AB=170t$ and $AO=340t$

In $△ABO$

$\cos \theta =\frac{170t}{340t}=\frac{1}{2}.......\left(1\right)$

Now by using Doppler effect, apparent frequency heard,

$f=\left(\frac{v+v_0}{v-v_s}\right)\times f_0$

where, $v_0=$speed of observer and $v_s=$speed of source (both speed should be along line joining the observer and source)

$\Rightarrow f=\left(\frac{v+0}{v-v_s\cos \theta }\right)\times f_0$

Substituting the given values and using $\left(1\right)$ ,

$f=\left(\frac{340+0}{340-170\times \frac{1}{2}}\right)\times 1200\Rightarrow f=1600\text{Hz}$

Hence, option (b) is the correct answer.