Question

A source emitting sound at frequency $3000\text{Hz}$, is moving along positive $y-$ axis with a speed $35\text{m/s}$. A listener is situated on the ground at the position $(670\text{m},0)$. Find the frequency of the sound received by the listener at the instant the source crosses the origin. Speed of sound in air is $335\text{m/s}$

• A. $3000\text{Hz}$
• B. $2568\text{Hz}$
• C. $3033\text{Hz}$
• D. $3350\text{Hz}$

Answer 1

The correct option is C $3033\text{Hz}$



Given that,
Speed of sound in stationary medium $\left(v\right)=335\text{m/s}$

Actual frequency $f_0=3000\text{Hz}$

Velocity of source of sound $\left(v_s\right)=35\text{m/s}$

At $t=0$, source is emitting sound and after time $t_1$ it crosses the origin.

This means sound also travelled from B to A in time $t_1$.

$\Rightarrow \frac{BC}{v_s}=\frac{AB}{v}$

$\Rightarrow \frac{BC}{AB}=\frac{v_s}{v}=\frac{35}{335}$

Now In $\Delta ABC$,

$\cos \theta =\frac{BC}{AB}=\frac{35}{335}$

From the definition of Doppler effect,

$f_{app}=f_0\left(\frac{v\pm v_o}{v\pm v_s}\right)$

Since the listener has fixed position, we can say that, $v_o=0$, and source is approching the listener along the line joining them.

$\Rightarrow f_{app}=f_0\left(\frac{v}{v-v_s\cos \theta }\right)$

Substituting the data given in the question,

$f_{app}=3000\times \left(\frac{335}{335-35\times \frac{35}{335}}\right)$
$=\frac{3000\times 335\times 335}{{335}^2-{35}^2}\approx 3033\text{Hz}$

Note:- No matter where this point is on $X-$ axis, frequency of sound heard by every point on the $X-$ axis is same when the source crosess the origin .