Question

An engine whistling at a constant frequency $n_0$ and moving with a constant velocity goes past a stationary observer. As the engine crosses him, the frequency of the sound heard by him changes by a factor f. The actual difference in the frequencies of sound heard by him before and after the engine crosses him is

• A. $\frac{1}{2}{n}_0(1-f^2)$
• B. $\frac{1}{2}{n}_0\left(\frac{1-f^2}{f}\right)$
• C. $n_0\left[\frac{1-f}{1+f}\right]$
• D. None of these

Answer 1

The correct option is B $\frac{1}{2}{n}_0\left(\frac{1-f^2}{f}\right)$

Let v be the velocity of the engine and V be the velocity of sound the frequency of the sound heard as the engine approaches to observer is given by $n_1=\frac{n_0}{1-\frac{v}{V}}$
The frequency of the sound heard as the engine recedes from the observer is given by $n_2=\frac{n_0}{1+\frac{v}{V}}$
It is given that $n_2=f{n}_1$
$\therefore \frac{n_0}{V+v}=f\frac{n_0}{V-v}$ or $V-v=fV+fv$
$\therefore V(1-f)=v(1+f)$ or $\frac{v}{V}=\frac{1-f}{1+f}$
The difference in the frequencies of the sound heard is
$n_1-n_2=n_1(1-f)=\left(\frac{n_0}{1-\frac{1-f}{1+f}}\right)(1-f)=\frac{n_0(1-f^2)}{2f}$