Question

At each of two stations $A$ and $B$, a siren is sounding with a constant frequency of $250\text{cycle}/s$. A cyclist from $A$ proceeds straight towards $B$ with a velocity of $12km/h$ and hears $5\text{beats}/s$. The velocity of sound is nearly:

• A. $336m/s$
• B. $320m/s$
• C. $328m/s$
• D. $333m/s$

Answer 1

The correct option is D $333m/s$

Given, two stations $A$ and $B$.
Actual frequency, $n=250\text{cycle}/s$
Velocity of source,
$v_s=12km/h=12\times \frac{5}{18}=3.33m/s$
Beat frequency, $n_B-n_A=5\text{beats}/s$
If $n_B$ is the frequency as heard by cyclist from source B.

$n_B=n\left(\frac{v_s+v_0}{v_s}\right)...\left(i\right)$

If $n_A$ is the frequency as heard by the cyclist from source $A$

$n_A=n\left(\frac{v_s-v_0}{v_s}\right)...\left(ii\right)$

Given, beat frequency,

$n_B-n_A=5\text{beats}/s$

From equation $\left(i\right)$ and $\left(ii\right)$,

$$\begin{array}{cc}{n}_B-n_A& =n\left(\frac{v_s+v_0}{v_s}\right)-n\left(\frac{v_s-v_0}{v_s}\right)\\ 5& =250\left(\frac{v_s+3.33}{v_s}\right)-250\left(\frac{v_s-3.33}{v_s}\right)\\ 5& =250[\left(\frac{v_s+3.33}{v_s}\right)-\left(\frac{v_s-3.33}{v_s}\right)]\\ & \\ 5& =250\left[\frac{v_s+3.33-v_s+3.33}{v_s}\right]\\ & \\ v_s& =\frac{250}{5}\left(6.66\right)\\ & \\ v_s& =333m/s\end{array}$$

Final answer: (c)