Question

The apparent frequency of the whistle of an engine changes by the ratio $\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$ as the engine passes a stationary observer. If the velocity of sound is $340m{s}^{-1}$, then the velocity of the engine is

Answer 1

Step 1. Given data:

Whistle variation ratio = $\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$

The velocity of sound $\left(v_s\right)$= $340m{s}^{-1}$

Let the frequency of the whistle be $n$ and the velocity of the engine be $v$.

Let the frequencies as heard by the stationary observer be $n^{\text{'}}$ and $n^{\text{'}\text{'}}$ before and after the source passes the stationary observer.

Step 2. Calculations:

Doppler effect when source $\left(S^{\text{'}}\right)$ approaches the stationary observer:

$n^{\text{'}}=n\left[\frac{v_{sound}}{v_{sound}-v_{source}}\right]=n\left[\frac{340}{340-v}\right]...........\left(1\right)$

Doppler effect when the source $\left(S^{\text{'}}\right)$ moves away from the stationary observer:

$n\text{'}\text{'}=n\left[\frac{v_{sound}}{v_{sound}+v_{source}}\right]=n\left[\frac{340}{340+v}\right]..........\left(2\right)$

Now $\frac{n\text{'}}{n\text{'}\text{'}}=\frac{9}{8}$ (Given)

Dividing $\left(1\right)$ by $\left(2\right)$, we get

$\Rightarrow \frac{340+v}{340-v}=\frac{9}{8}$

After simplification.

$v=20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Thus, the velocity of the engine is $20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.