Question

Two cars $A$ and $B$ are moving on two separate perpendicular roads approaching a crossing with uniform speed of $72\text{km/h}$ and $36\text{km/h}$ respectively. If the car $B$ blows horn of frequency $280\text{Hz}$ then the frequency of horn heard by the driver of the car $A$, when the line joining the cars make an angle of ${45}^{\circ }$ with the roads, will be,
[speed of sound $=330\text{m/s}$]

• A. $321\text{Hz}$
• B. $298\text{Hz}$
• C. $289\text{Hz}$
• D. $280\text{Hz}$

Answer 1

The correct option is B $298\text{Hz}$

Speed of car $A$, $v_A=72\text{km/hr}=20\text{m/s}$
Speed of car $B$, $v_B=36\text{km/hr}=10\text{m/s}$

According to Doppler's effect, the apparent frequency heard by the observer is,
$f^{{}^{\prime }}=f_0\left(\frac{v\pm v_0}{v\pm v_s}\right)$

The frequency heard by observer,
$f^{{}^{\prime }}=f_0\left(\frac{v+v_0}{v-v_s}\right)$
[as separation is decreasing, so apparent frequency will be greater than actual frequency]


$\Rightarrow f^{{}^{\prime }}=f_0\left(\frac{v+v_A\cos {45}^{\circ }}{v-v_B\cos {45}^{\circ }}\right)$

[we take the components of, velocities of the source and observer, along the line joining the source and observer]

$\Rightarrow f^{{}^{\prime }}=280\times \left(\frac{330+20\times \frac{1}{\sqrt{2}}}{330-10\times \frac{1}{\sqrt{2}}}\right)$

$\therefore f^{{}^{\prime }}=298\text{Hz}$

Hence, $\left(B\right)$ is the correct answer.