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Question: A source of sound is travelling at \(\dfrac{{100}}{3}m{s^{ - 1}}\) along a road, towards a point \(A...

A source of sound is travelling at 1003ms1\dfrac{{100}}{3}m{s^{ - 1}} along a road, towards a point AA. When the source is 3m3m away from AA, a person standing at a point OO on a road perpendicular to the track hears a sound of frequency ff. The distance of OO from AA at that time is 4m4m. If the original frequency is 640Hz640Hz, then the value of ff' is (given velocity of sound =340ms1 = 340m{s^{ - 1}})
A. 620Hz620Hz
B. 680Hz680Hz
C. 720Hz720Hz
D. 840Hz840Hz

Explanation

Solution

In order to find the solution of the given question, first of all we need to find the component of the velocity in which it is acting. After that we need to apply the formula for the frequency according to the Doppler’s effect. Then we need to solve the equation obtained and finally we can conclude with the correct solution for the given question.

Complete step by step answer:
Step 1:
First of all let us draw a figure according to the given conditions.

As we can see from the figure that the situation is of a right angle triangle, so we can find the hypotenuse which will be, h=32+42=25=5mh = \sqrt {{3^2} + {4^2}} = \sqrt {25} = 5m
Now, in the triangle, cosθ=bh=35\cos \theta = \dfrac{b}{h} = \dfrac{3}{5}

Step 2:
Now, we need to find the horizontal component of the source of the sound as the sound is travelling along a road. So, we can write, vs=vcosθ=1003×35=20ms1{v_s} = v\cos \theta = \dfrac{{100}}{3} \times \dfrac{3}{5} = 20m{s^{ - 1}}

Step 3:
We know the formula for frequency in case of Doppler’s effect is given by,
f=f(v±vov±vs)f' = f\left( {\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}}} \right) -----(i)
Here, since the observer is not moving so the velocity of the observer is zero.
Now, putting the values in equation (i), we get,
f=640(340034020)f' = 640\left( {\dfrac{{340 - 0}}{{340 - 20}}} \right)
f=640×340320=680Hz\Rightarrow f' = 640 \times \dfrac{{340}}{{320}} = 680Hz
Therefore, the required value of the frequency is 680Hz680Hz.

Hence, the correct answer is option (B).

Note: We define Doppler’s effect as the effect observed when either the source of a wave moves or the observer moves relative to each other. In other words we can say that the increase or decrease in the frequency of a wave when the source of the wave and the observer move towards or away from each other.