Question

Consider a source of sound S and an observer P. The sound source is of frequency $n_0.$ The frequency observed by P is found to be $n_1$ if P approaches S at speed v and S is stationary ; $n_2$ if S approaches P at a speed v and P is stationary and, $n_3$ if each of P and S has speed v/2 towards one another. Now,

• A. $n_1=n_2=n_3$
• B. $n_1<n_2$
• C. $n_3>n_0$
• D. $n_3$ lies between $n_{1}and{n}_2$

Answer 1

Answer: (B), (C), (D)

$n_3$ lies between $n_{1}and{n}_2$

When observer P approaches the the stationary source at speed v,
$n_1=\frac{V+v}{V}\times n_0$ (i)
($V$ is speed of sound)
When source S approaches the stationary observer P at speed v,
$n_2=\frac{V}{V-v}\times n_0$ (ii)
Thus, $n_2>n_1,$ i.e., choice (b) is correct when both S and P approach each other with speed v/2
$n_3=\frac{V+\left(v/2\right)}{V-\left(v/2\right)}{n}_0$ (iii)
Hence, $n_3>n_{0}and{n}_3$ lies between $n_{1}and{n}_2.$

Why this question? Concepts: It is a classic example which covers the cases of source moving, observer moving and both moving simultaneously, and also displays the power of multiple variants of one single formula!