# Question #a6987

Aug 8, 2017

when the speeds of source and the receiver relative to the air are lower than the velocity of sound in the air, the relationship between observed frequency $\nu$ and frequency of sound emitting from source ${\nu}_{0}$ is given by Doppler's effect as follows

$\nu = \frac{v + {v}_{r}}{v + {v}_{s}} {\nu}_{0.} \ldots \left[1\right]$,

where

• $v$ is the velocity of sound in the air;

• ${v}_{r}$ is the velocity of the receiver of sound relative to the air. It is positive if the receiver is moving towards the source and negative in the other direction;

• ${v}_{s}$ is the velocity of the source relative to the air. It is positive if the source is moving away from the receiver and negative in the reverse direction.

• Let at an instant the line joining the position of 1st car ($A$) and position of the 2nd car ($B$) makes an angle $\theta$ with the direction of motion of the first car. So this line will make angle ${\left(90 - \theta\right)}^{\circ}$ with the direction of motion of the 2nd car.

• So the component of velocity of the 1st car along $A B$ will be ${v}_{A B} = {v}_{1} \cos \theta$,where ${v}_{1}$ is the velocity of the 1st car towards the crossing.

• And the component of velocity of the 2nd car along $B A$ will be ${v}_{B A} = v \cos \left(90 - \theta\right) = {v}_{2} \sin \theta$, where ${v}_{2}$ is the velocity of the 2nd car towards the crossing.

So

${v}_{r} = {v}_{2} \sin \theta$

and ${v}_{s} = - {v}_{1} \cos \theta$

Now the equation [1] becomes

$\nu = \left(\frac{v + {v}_{2} \sin \theta}{v - {v}_{1} \cos \theta}\right) {\nu}_{0}$

Now

• ${v}_{1} = 72 k m \text{/"hr=20m"/} s$

• ${v}_{2} = 36 k m \text{/"hr=10m"/} s$

• $v = = 340 m \text{/} s$

• ${\nu}_{0} = 280 H z$

So

$\nu = \left(\frac{340 + 10 \sin \theta}{340 - 20 \cos \theta}\right) 280$ Hz

When $\theta = {45}^{\circ}$ we finally get the frequency of horn heard by the driver of 2nd car.

$\nu = \left(\frac{340 + 10 \sin 45}{340 - 20 \cos 45}\right) 280 \approx 298$ Hz