Question

The velocity of sound in air is \[330m{s^{ - 1}}\]. To increase the apparent frequency of the sound by \[50\% ,\], the source should move towards the stationary source with a velocity equal to
\[
  A.\;\;\;\;\;110{\text{ }}m{s^{ - 1}} \\
  B.\;\;\;\;\;105{\text{ }}m{s^{ - 1}} \\
  C.\;\;\;\;\;220{\text{ }}m{s^{ - 1}} \\
  D.\;\;\;\;\;330{\text{ }}m{s^{ - 1}} \\
  \]

Answer 1

Hint: Use the Doppler Effect concept. Note the direction in which the source is moving and find the ratio of frequency by the condition. Now, substitute the value in terms of frequency of source and simplify to find the value of \[\;{v_s}\].

Complete step-by-step solution:
Whenever there is a relative motion between a source and the observer, the frequency of sound heard by the observer is different from the actual frequency of sound emitted by the source. The frequency observed by the observer is called apparent frequency. It can be more or lesser than the actual frequency which depends on the relative motion.
In this case the source is moving towards the observer, so apparent frequency =
$n' = n\left( {\dfrac{v}{{v - {v_s}}}} \right)$
Where,
n=frequency of source
v=velocity of sound
\[\;{v_s}\] = velocity of source
According to the problem,
$
  n' = n + \dfrac{{50}}{{100}}n \\
  n' = \dfrac{3}{2}n \\
$
Substituting the value in the expression, we get
$
  \dfrac{3}{2}n = n\left( {\dfrac{v}{{v - {v_s}}}} \right) \\
  3v - 3{v_s} = 2v \\
  {v_s} = \dfrac{v}{3} \\
  {v_s} = \dfrac{{330}}{3} \\
  {v_s} = 110m{s^{ - 1}} \\
  $
So, to increase the apparent frequency by 50% the velocity of source should be 110m/s.

A is the correct option.

Note: Some conventions:
1. The direction of v is always taken from source to observer
2. All the velocities in the direction of v are taken positive
3. All the velocities in the opposite direction of v are taken negative.