Question

A train moves towards a stationary observer with speed $34m/s$. The train sounds a whistle and its frequency registered by the observer is ${f_{(1)}}$. If the train's speed is reduced to $17m/s$ the frequency registered is ${f_{(2)}}$ . If the speed of sound of $340m/s$ then the ratio ${f_{(1)}}/{f_{(2)}}$ is.
A). 18/17
B). 19/18
C). 20/19
D). 21/20

Answer 1

Hint: Doppler effect can be considered as a property of sound wave. It is a change in sound wave frequency caused by the relative movement of either source, listener or both. Use the equation of Doppler effect to find the respective frequencies heard by the listener when the train is travelling at 34m/s and 17m/s and take their ratio.

Complete step by step answer:
When a source of sound approaches us ,the pitch becomes higher and as it moves away from the listener ,pitch gets lower. It is due to the change in the number of waves passing through per unit area per unit time ,i.e. due to the frequency. As someone approaches a stationary listener the sound waves compress which increases the frequency resulting in higher pitch .Similarly as the source moves away , the number of waves decreases as it spread apart therefore frequency will decrease giving a lower pitch .Similarly when source and listener approach each other the frequency increases and as they move apart the frequency decreases.
Considering the influencing factors such as the speed of listener or source, either it is approaching or going further away the equation of Doppler effect can be written as
${f_L} = \dfrac{{v + {v_l}}}{{v + {v_s}}}{f_s}$
Here${f_L}$ is the frequency that the listener hears and ${f_s}$ is the frequency of the sound
$v$ is the velocity of sound in the medium it is travelling in m/s,
${v_l}$ is the velocity of the listener in m/s and ${v_s}$ is the velocity of source in m/s.
In the given question ,the value of ${v_l}$ is zero as the listener is stationary for both the cases. Similarly $v$ is $340m/s$
First case
${v_s}$ =$34m/s$
${f_1} = \dfrac{{340 + 0}}{{340 - 34}}{f_s}$
${f_1} = \dfrac{{10}}{9}{f_S}$
Second case
${v_s}$ =$17m/s$
${f_2} = \dfrac{{340 + 0}}{{340 - 17}}{f_s}$
${f_2} = \dfrac{{20}}{{19}}{f_S}$
Taking ratio of both the cases we get
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\dfrac{{10}}{9}}}{{\dfrac{{20}}{{19}}}}$
$\dfrac{{{f_1}}}{{{f_2}}}$$ = \dfrac{{19}}{{18}}$
Option A is correct.

Note: Similarly Doppler effect is also seen in light as well. Due to that any light moving towards the observer will be shifted to lower wavelengths .This is called Blue shift. Similarly light sources moving far away from the observer will shift to longer wavelengths .This is called red shift. Since the speed of all electromagnetic Waves are relatively very large compared to the speed of source or listener in our day to day life , the apparent change in frequency is not easily noticed . The apparent frequency is almost equal to its original frequency.(The ratio $\dfrac{{v + {v_l}}}{{v + {v_s}}}$ is nearly one for light in the equation ${f_L} = \dfrac{{v + {v_l}}}{{v + {v_s}}}{f_s}$.)