Solveeit Logo

Question

Question: A rocket is moving at a speed of \(200\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}\) towards a stationar...

A rocket is moving at a speed of 200  ms - 1200\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}} towards a stationary target. While moving, it emits a wave of frequency 1000  Hz1000\;{\text{Hz}} Calculate the frequency of the sound as detected by the target. (Velocity of sound in air is 330  ms - 1330\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}})

Explanation

Solution

Doppler effect used to find the frequency of the sound detected from the target. The Doppler effect or Doppler shift is used to establish a relation between the observer's velocity towards the source and source velocity towards the observer.

Useful formula:
The expression for finding the frequency of sound wave detected from the target is
f=f0(vsoundvsoundvs)f = {f_0}\left( {\dfrac{{{v_{sound}}}}{{{v_{sound}} - {v_s}}}} \right)
Where vs{v_s}is the speed of the rocket, vsound{v_{sound}}is the velocity of sound in air and f0{f_0}is the original sound frequency.

Complete step by step answer:
Given, The speed of the rocket is vs=200  ms - 1{v_s} = 200\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}
Original sound frequency is fo=1000  Hz{f_o} = 1000\;{\text{Hz}}
Velocity of sound in air is Vsound=330  ms - 1{V_{sound}} = 330\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}
we need to find the frequency of sound as detected by the target.

The expression for finding the frequency of sound wave detected from the target is

f=f0(vsoundvsoundvs)f = {f_0}\left( {\dfrac{{{v_{sound}}}}{{{v_{sound}} - {v_s}}}} \right)
Substitute all the value in the above equation
 f=1000  Hz(330  ms - 1330  ms - 1200  ms - 1) f=1000  Hz(330  ms - 1130  ms - 1) f=1000  Hz×2.54   f=2540  Hz \ f = 1000\;{\text{Hz}}\left( {\dfrac{{330\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}}}{{330\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}} - 200\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}}}} \right) \\\ f = 1000\;{\text{Hz}}\left( {\dfrac{{330\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}}}{{130\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}}}} \right) \\\ f = 1000\;{\text{Hz}} \times 2.54\; \\\ f = 2540\;{\text{Hz}} \\\

Thus, the frequency of sound wave detected from target is f=2540  Hzf = 2540\;{\text{Hz}}

Additional information:
Doppler effect states, an increase or decrease frequency of light is obtained as the source and observer move toward or away from each other. The combined equation is used when both are moving relative to one and another. The application of Doppler effects is- radar, vibration measurement, medical imaging, etc.

Note:
By using the data of frequency of sound wave detected from the target, velocity of the sound and velocity of the rocket the frequency of echo as detected by the rocket can be calculated by using the following formula.
fecho=f(vsound+vsvsound){f_{echo}} = f\left( {\dfrac{{{v_{sound}} + {v_s}}}{{{v_{sound}}}}} \right)
Where vs{v_s}is the speed of the rocket, vsound{v_{sound}}is the velocity of sound in air and ffis the frequency of sound wave detected from target.