Question

A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f₁. If the speed of the train is reduced to 17 m/s, the frequency registered is f₂. If speed of sound is 340 m/s, then the ratio f₁/f₂ is :
(A) 18/17 (B) 19/18
(C) 20/19 (D) 21/20

Answer 1

Given:

• Speed of train $(V_t) = 34 m/s$
• Frequency of whistle should  $= F_1$
• Speed of the train is reduced $= 17 m/s$
• Speed of sound $(V) = 340 m/s$
• Frequency of sound after reducing train speed$= F_2$

To Find:

• Find the ratio of frequency  $F_1$ and  $F_2$ .

Solution:

Let f is the actual frequency

We know that,

When train approaches to the observer, then apparent frequency is given by:                

                $F_1 = (\frac{V}{V- V_t} ) \times F$

      Putting all the values

We get,

             $F_1 = (\frac{340}{340- 34} ) \times F$

             $F_1 = (\frac{340}{306} ) \times F$      $.........................Eq(i)$

And,

            $F_2 = (\frac{V}{V-V_t} ) \times F$

Here,               $V_t =$ Reduced speed $= 17m/s$

So,

                     $F_2 = (\frac{340}{340-17} ) \times F$

                     $F_2 = (\frac{340}{323} ) \times F$    $..........................eq(ii)$

From $eq(i)$ and $eq(ii)$

                            $\frac{F_1}{F_2} = \frac{340}{306} \times \frac{323}{340}$

                            $\frac{F_1}{F_2} = \frac{323}{306}$

                           $\frac{F_1}{F_2} = \frac{19}{18}$

Thus, The ratio of frequency of  $F_1$ and  $F_2$   is $19:18$.