Question

The apparent frequency of the whistle of an engine changes by the ratio \[\dfrac{5}{3}\] as the engine passes a stationary observer. If the velocity of sound is 340 \[m{s^{ - 1}}\], then the velocity of the engine is:
A.340 \[m{s^{ - 1}}\]
B.170 \[m{s^{ - 1}}\]
C.85 \[m{s^{ - 1}}\]
D.42.5 \[m{s^{ - 1}}\]

Answer 1

Hint: This question is based on the doppler effect of sound. Doppler effect of sound is an increase (or decrease) in the frequency of sound as the source and the observer move towards (or away from) each other.

Complete answer:
The Doppler effect is the change in the frequency of a wave observed by an observer who is moving relative to the wave source. This phenomenon was described by Christian Doppler in 1842.
Doppler effect states that the apparent frequency to the observer is equal to:
\[{f_{app}} = f(\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}})\]
where \[{f_{app}}\] is the apparent frequency, f is the natural frequency, \[{v_o}\] is the velocity of the observer and \[{v_s}\] is the velocity of the source.
Hence,
\[{f_1} = f(\dfrac{{v \pm (0)}}{{v - {v_s}}})\]
And \[{f_2} = f(\dfrac{{v \pm (0)}}{{v + {v_s}}})\]
We know that,
v = 340 \[m{s^{ - 1}}\]
Velocity of the source (), that is the velocity of the engine is unknown.
Now, given that \[\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{5}{3}\]
\[ \Rightarrow \dfrac{{f(\dfrac{{v \pm (0)}}{{v - {v_s}}})}}{{f(\dfrac{{v \pm (0)}}{{v + {v_s}}})}} = \dfrac{5}{3}\]
\[ \Rightarrow \dfrac{{v + {v_s}}}{{v - {v_s}}} = \dfrac{5}{3}\]
 3v + 3 \[{{v_s}}\] = 5v - 5 \[{{v_s}}\]
 8 \[{{v_s}}\] = 2v
 \[{{v_s}}\] = \[\dfrac{v}{4}\]
 \[{{v_s}}\] = \[\dfrac{340}{4}\]
 \[{{v_s}}\] = 85 \[m{s^{ - 1}}\]
Hence, the velocity of source, i.e. the velocity of engine is equal to 85 \[m{s^{ - 1}}\]

Therefore, the correct answer is: (C) 85 \[m{s^{ - 1}}\] .

Note:
The sign of the source velocity and the object velocity in the doppler formula depends on whether their distance is being increased or decreased. Remember that the velocity vector which decreases the distance between the two increases the frequency and vice versa.