Solveeit Logo
Login

Question

Question: A car fitted with a device which transmits sound \(60\) times per minute. There is no wind and speed...

A car fitted with a device which transmits sound 6060 times per minute. There is no wind and speed of sound in still air is 345msec⁡−1345m{\sec ^{ - 1}} if you hear the sound 6868 times per minute when you are moving towards the car with a speed of 12msec⁡−112m{\sec ^{ - 1}} the speed of car must be nearly.
[A] 20msec⁡−120m{\sec ^{ - 1}} Towards you
[B] 30msec⁡−130m{\sec ^{ - 1}} Towards you
[C] 10msec⁡−110m{\sec ^{ - 1}} Away from you
[D] 10msec⁡−110m{\sec ^{ - 1}} Towards you

Explanation

Solution

In order to solve this question we need to understand the Doppler Effect of sound. Since we know sound waves are longitudinal waves in nature and they propagate with compression and rarefaction, sound waves also need a medium to travel. Doppler Effect states there is relative motion between observer and listener then due to relative motion frequency heard by observer would be different. A real life example of this is when you stand at train stations then you hear engine sound from very far away but it is of very less intensity but as engine moves toward you find there is increase in its frequency, this is Doppler Effect.

Complete step-by-step solution:
Given, frequency of source is, f0=60{f_0} = 60 per minute or f0=1{f_0} = 1 per second or f0=1Hz{f_0} = 1Hz
Frequency that observer listens, f=68f = 68 per minute or f1=1.13{f_1} = 1.13 per second or f1=1.13Hz{f_1} = 1.13Hz
Let the listener moving toward source with velocity vr{v_r} given as, vr=12msec⁡−1{v_r} = 12m{\sec ^{ - 1}}
And sound speed in medium is, v=345msec⁡−1v = 345m{\sec ^{ - 1}}
Let the source moving towards observer with speed vo{v_o}
Then according to Doppler Effect frequency of wave which listener listens is given by, f=(v+vrv−v0)f0f = (\dfrac{{v + {v_r}}}{{v - {v_0}}}){f_0}
So speed of source is, ff0=v+vrv−v0\dfrac{f}{{{f_0}}} = \dfrac{{v + {v_r}}}{{v - {v_0}}}
v−v0=(v+vrf)f0v - {v_0} = (\dfrac{{v + {v_r}}}{f}){f_0}
vo=v−(v+vrf)f0{v_o} = v - (\dfrac{{v + {v_r}}}{f}){f_0}
Putting values we get, v0=345−(345+121.13)1{v_0} = 345 - (\dfrac{{345 + 12}}{{1.13}})1
v0=345−315.92{v_0} = 345 - 315.92
v0=29.08msec⁡−1{v_0} = 29.08m{\sec ^{ - 1}}
So the correct option is, [B] 30msec⁡−130m{\sec ^{ - 1}} towards you

Note: It should be remembered that speed of sound depends on the medium in which it propagates because it is a longitudinal wave while speed of light does not depend on the medium to travel as it is a transverse wave. Longitudinal waves are those waves whose displacement is not perpendicular to wave propagation while in transverse wave displacement is perpendicular to wave propagation.