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Question

A train blowing its whistle moves with a constant velocity v away from an observer on the ground. The ratio of the natural frequency of the whistle to that measured by the observer is found to be 1.2. If the train is at rest and the observer moves away from it at the same velocity, this ratio would be given by


A
0.51
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B
1.25
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C
1.52
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D
2.05
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Solution

The correct option is B 1.25

If the train is going away from the observer, the apparent frequency is

$$v_1=\frac{vu}{v+u}=\frac{v}{1+ \frac{u}{v}}$$            ..........(1)

It is observed that $$v=1.2\, v_1$$ (Given),

In the second case the apparent frequency is 

$$v_2=\frac{v(v-u)}{v}=v(1-\frac{u}{v})$$

or

$$\frac{v}{v_2}=\frac{1}{1-\frac{u}{v}}$$                     ............(2)

Now, from equation (1) we have 

$$\frac{v}{v_1}=1+\frac{u}{v}$$

or

$$1.2=1+\frac{u}{v}$$

$$u=0.2v$$

That is,

$$\frac{u}{v}=0.2$$

Using this in equation (2), we get,

$$\frac{v}{v_2}=\frac54=1.25$$

Physics

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