Question

A person in topless jeep is travelling with uniform velocity $20{\text{ }}m{s^{ - 1}}$away from a vertical mountain. The person blows the horn of the jeep and hears its echo from the mountain. If actual frequency of horn is 630 Hz, the apparent decrease in the frequency of echo with respect to actual frequency of the horn is: (velocity of sound in air = $340{\text{ m}}{{\text{s}}^{ - 1}}$ )
A) 35 Hz
B) 45 Hz
C) 70 Hz
D) 90 Hz

Answer 1

In this question we have to find the decrease in apparent frequency echo with respect to actual frequency of the horn. For this first we are going to use the formula of apparent frequency. Then we are going to use the frequency of echo.

Complete step by step solution:
Given,
The velocity of the person ${v_0 } = {v_S} = 20m{s^{ - 1}}$
Where,
${v_0 }$ is the velocity of the observer and ${v_S}$ is the velocity of source.
The formula of apparent velocity for a moving away is given by following formula,
${\text{Apparent frequency }}{{\text{f’}}} = \left( {\dfrac{v}{{v + {v_S}}}} \right)f$…………(1)
${{\text{f’}}}$ is the frequency heard by the person and ${\text{f}}$ is the frequency of the horn.
$Frequency{\text{ }}of{\text{ }}echo{\text{ }}{f’’} = \left( {\dfrac{{v - {v_0 }}}{{v + {v_S}}}} \right)f$………..(2)
Putting values of v, f and ${v_S}$in equation (1)
$\Rightarrow {\text{Apparent frequency }}{{\text{f’}}} = \left( {\dfrac{{340}}{{340 + 20}}} \right)630$
$\Rightarrow {\text{Apparent frequency }}{{\text{f’}}} = \left( {\dfrac{{340}}{{360}}} \right)630$
$\Rightarrow {\text{Apparent frequency }}{{\text{f’}}} = \left( {\dfrac{{2142}}{{360}}} \right)$
$\Rightarrow {\text{Apparent frequency }}{{\text{f’}}} = 595Hz$
Now, to find the value of frequency of echo we will put the values of all the variables in equation (2)
$\Rightarrow Frequency{\text{ }}of{\text{ }}echo{\text{ }}{f''} = \left( {\dfrac{{340 - 20}}{{340 + 20}}} \right)360$
$\Rightarrow Frequency{\text{ }}of{\text{ }}echo{\text{ }}{f’'} = \left( {\dfrac{{320}}{{360}}} \right)360$
$\Rightarrow Frequency{\text{ }}of{\text{ }}echo{\text{ }}{f’’} = \left( {\dfrac{{201600}}{{360}}} \right)$
$\Rightarrow Frequency{\text{ }}of{\text{ }}echo{\text{ }}{f''} = 560Hz$
The apparent change in frequency = (630-560)Hz
The apparent change in frequency =70 Hz
Result- Hence, from above frequency the apparent change in the frequency of echo is 70 Hz.

Hence, option (C) is correct.

Note: In this question we have used the formulae of frequency of echo and observer; so the knowledge of these formulae should be good. We should be careful while finding the values of different variables as it matters a lot otherwise we will get the wrong answer. We should be careful while doing the calculations. It is very important to understand which value will be the velocity of an observer and which one will be the velocity of source. And we can remember this that v is always the velocity of sound in case of problems including sound and echo. In relative motion v is always the velocity of the reference frame.