Doppler Effect: Definition, Formula and Application

Doppler Effect: Have you ever wondered that when a whistling train is approaching a railway crossing, the pitch of its whistle keeps on increasing, and when it recedes away, the pitch decreases. How does the radar gun estimate the velocity of a moving vehicle, or how is the velocity of a star far away in the universe estimated? What do we mean when we say that Blue shift or Red shift?

All these seemingly disconnected things are in fact connected by an amazing behaviour of waves known as the Doppler Effect. It is observed in both sound and electromagnetic waves. Let us try to understand this phenomena with the help of this article.

What is Doppler Effect?

Doppler Effect is the phenomenon in which the observed frequency of a wave is different from frequency of the source due to relative motion between the source and the observer. The observed frequency can be less or more than the source frequency depending upon the direction and magnitudes of the velocities of both the source and the observer.

For example: When we wait at the railway crossing, and the train crosses us, the sound of the siren changes its pitch as it crosses us. This happens because there is the relative velocity between the source and observer and while it is approaching, the relative velocity is negative and when it is moving away, the relative velocity is positive. Thus, there is a difference in the pitch of the sound while the train is approaching and while it is moving away.

Doppler Effect Derivation

Let us consider a source moving with a velocity \(‘{v_S}’\) and observer moving with the velocity \(‘{v_0}’.\)

The velocity of sound wave be \(‘v’.\)

The observed frequency be \(‘f’.\)

The source frequency be \(‘{f_0}’.\)

Time period of the source \(‘{T_0}’.\)

Time period observed by the observer \(‘{T_{{\rm{obs}}}}’\)

The initial distance between the source and the observer be \(‘d’\) The initial time instance is \(‘0’\)

Let the time instance at which the first instance of the wave reaches the observer be \(‘{t_1}’\)

\(v{t_1} + {v_o}{t_1} = d\)

The time instance at which the source will produce the same wave will be after its time period \(‘{T_0}’\)

The sound produced at the time instance, \(t = {T_0},\) reaches the observer at the time instance \(t = {t_2}\)

\(\left( {{t_2} – {T_0}} \right)v = d – {v_s}{T_0} – {v_0}{t_2}\)

On solving we get,

\({t_2} = \frac{{d – {v_s}{T_0} – v{T_0}}}{{v + {v_0}}}\)

One complete cycle of the sound wave is produced by the source in \({T_0} – 0 = {T_0},\) i.e., the time period of the source.

But, the observer observes the same sound wave in \({t_2} – {t_1} = {T_{{\rm{obs}}}},\) i.e., the observed time period.

Thus, on solving we get,

\({T_{{\rm{obs}}}} = {t_2} – {t_1}\)

\( \Rightarrow {T_{{\rm{obs}}}} = \frac{{d – {v_s}{T_0} + v{T_0}}}{{v + {v_0}}} – \frac{d}{{v + {v_0}}}\)

\( \Rightarrow {T_{{\rm{obs}}}} = \left( {\frac{{v – {v_s}}}{{v + {v_0}}}} \right){T_0}\)

Since the frequency equal to the reciprocal of the time period we have, \(f = {f_0}\frac{{\left( {v + {v_0}} \right)}}{{\left( {v – {v_s}} \right)}}\)

Doppler Effect in Sound Wave

Velocity of Observer	Velocity of Source	Apparent Frequency
Towards the source. Velocity of observer is positive.	Towards the observer. Velocity of source is positive.	\(f = {f_0}\frac{{\left( {v + {v_0}} \right)}}{{\left( {v – {v_s}} \right)}}\)
Towards the source. Velocity of observer is positive.	Away from observer. Velocity of source is negative.	\(f = {f_0}\frac{{\left( {v + {v_0}} \right)}}{{\left( {v – \left( { – {v_s}} \right)} \right)}}\) \( \Rightarrow f = {f_0}\frac{{\left( {v + {v_0}} \right)}}{{\left( {v + {v_s}} \right)}}\)
Away from source. Velocity of observer is negative.	Towards the observer. Velocity of source is positive.	\(f = {f_0}\frac{{\left( {v + \left( { – {v_0}} \right)} \right)}}{{\left( {v – {v_s}} \right)}}\) \( \Rightarrow f = {f_0}\frac{{\left( {v – {v_0}} \right)}}{{\left( {v – {v_s}} \right)}}\)
Away from source. Velocity of observer is negative.	Away from observer. Velocity of source is negative.	\(f = {f_0}\frac{{\left( {v + \left( { – {v_0}} \right)} \right)}}{{\left( {v – \left( { – {v_s}} \right)} \right)}}\) \( \Rightarrow f = {f_0}\frac{{\left( {v – {v_0}} \right)}}{{\left( {v + {v_s}} \right)}}\)

Doppler Effect When Velocities are in Different Direction

Only the component of velocities in the direction of the line joining the source and the observer will be taken into consideration while applying the Doppler effect. In other words, the velocity component, which is perpendicular to the line joining the source and the observer does not affect the observed frequency.

Thus, the equation for Doppler effect becomes,

\(f = {f_0}\left( {\frac{{v + {v_0}{\rm{cos}}\left( \beta  \right)}}{{v – {v_s}{\rm{cos}}\left( \alpha  \right)}}} \right)\)

Where,

\({\rm{‘\alpha }}’\) is the angle between the velocity of the source and the line joining the source and the observer. \({\rm{‘\beta }}’\) is the angle between the velocity of the observer and the line joining the source and the observer.

Doppler Effect in the Presence of Wind

When the velocity of the medium is taken into consideration, we observe that the velocity of the wave itself gets changed.

If the velocity of the medium is from the source towards the observer, then the new velocity of the wave is the sum of the original velocity of the wave and the velocity of the medium. If the velocity of the medium is from the observer towards the source, then the new velocity of the wave is the difference of the original velocity of the wave and the velocity of the medium.

Velocity of the wind, \({v_{\rm{w}}}\)	Observed or Apparent frequency.
Towards the observer	\(f = {f_0}\left( {\frac{{\left( {v + {v_{\rm{w}}}} \right) + {v_0}}}{{\left( {v + {v_{\rm{w}}}} \right) – {v_s}}}} \right)\)
Towards the source	\(f = {f_0}\left( {\frac{{\left( {v – {v_{\rm{w}}}} \right) + {v_0}}}{{\left( {v – {v_{\rm{w}}}} \right) – {v_s}}}} \right)\)

Doppler Effect in Light Wave

The Doppler effect is applicable for all waves, including electromagnetic waves. Therefore, it is also applicable for lightwave but as the light wave has a very high speed of propagation and the white light is the combination of many waves, thus the Doppler effect is difficult to perceive with our naked eyes in case of light waves.

The apparent frequency is given by,

\(f = {f_0}\sqrt {\frac{{1 – \frac{v}{c}}}{{1 + \frac{v}{c}}}} \)

Where,

\(v\) is the velocity of source. \(c\) is the velocity of light.

Doppler Shift

The wavelength of light also changes and is given by,

\({\lambda _{{\rm{obs}}}} = {\lambda _s}\sqrt {\frac{{1 + \frac{v}{c}}}{{1 – \frac{v}{c}}}} \)

Where,

\({\lambda _s}\) is the emitted wavelength.

\({\lambda _{{\rm{obs}}}}\) is the observed wavelength.

When the observed wavelength of the light is longer. This is known as Red Shift, i.e., the observed light is more towards red in the light spectrum. When the observed wavelength of the light is shorter. This is known as Blue Shift, i.e., the observed light is more towards blue or violet in the light spectrum.

Application of Doppler Effect

1. Doppler effect is used to measure the speed of the motor car.

2. It is used to estimate the velocity of the galaxy and star.
3. It is used in the measurement of the Blood flow rate.

Summary

• 1. Doppler effect occurs due to relative motion between the source and observer.
• 2. Doppler effect occurs in sound wave as well as light wave.
• 3. Due to the doppler effect the wavelength of the light ray shifts in either the red side of the visible light spectrum also known as Red Shift or to the blue side of the visible light spectrum also known as Blue Shift. Collectively, this phenomenon is also known as Doppler Shift.
• 4. Apparent frequency is given by, \(f = {f_0}\frac{{\left( {v + {v_0}} \right)}}{{\left( {v – {v_s}} \right)}}\;\)
• Where,
  • The velocity of sound wave is \(‘v’.\)
  • The observed frequency is \(‘f’.\)
  • The source frequency is \(‘{f_0}’.\)
  • The source is moving with a velocity \(‘{v_S}’\) and observer is moving with the velocity \(‘{v_0}’.\)
• 5. Observer velocity towards source is positive and away from source negative.
• 6. Source velocity towards observer is positive and away for observer is negative.

Solved Questions on Doppler Effect

Q.1. Two cars are moving towards each other at the speed \({\rm{100\;m\;}}{{\rm{s}}^{{\rm{ – 1}}}}.\) If one of the cars honks with the frequency of \({\rm{300\;Hz}}{\rm{.}}\) What is the observed frequency? Also, find the perceived wavelength. Assume the speed of sound to be \({\rm{300\;m\;}}{{\rm{s}}^{{\rm{ – 1}}}}{\rm{.}}\)
Sol: Velocity of the cars is \({\rm{100\;m\;}}{{\rm{s}}^{{\rm{ – 1}}}}{\rm{.}}\)

Since the observer and the source both are moving towards each other,

Therefore, the apparent frequency will be given by, \(f = {f_0}\frac{{\left( {v + {v_0}} \right)}}{{\left( {v – {v_s}} \right)}},\)

Putting in the values, we get,

\(f = 300\frac{{\left( {300 + 100} \right)}}{{\left( {300 – 100} \right)}}\)

\( \Rightarrow f = 300 \times \frac{{400}}{{200}}\)

\( \Rightarrow f = 600\;{\rm{Hz}}\)

We know that,

\(v = f\lambda \)

Where,

\(v\) is the velocity of sound.

\(\lambda \) is the wavelength.

\(f\) is the frequency.

Putting in the value of the observed parameters to get the observed frequency,

\(300 = 600 \times \lambda \) \(\lambda  = 0.5\;{\rm{m}}\)

Q.2. A traffic policeman was standing on a road and sounds of whistle emitting the main frequency of \({\rm{2}}{\rm{.00\;kHz}}\) What could be the apparent frequency heard by or scooter driver approaching the policeman at a speed of \({\rm{36}}{\rm{.0\;km\;}}{{\rm{h}}^{{\rm{ – 1}}}}{\rm{?}}\) The speed of the sound in air is.
Sol: Let the speed of the scooter driver be \({v_0} = 36\;{\rm{km\;}}{{\rm{h}}^{{\rm{ – 1}}}} = {\rm{10\;m\;}}{{\rm{s}}^{{\rm{ – 1}}}}.\)

\(v\) is the velocity of sound.

\(f\) is the observed frequency.

\({f_0}\) is the emitted frequency.

The apparent frequency will be given by, \(f = {f_0}\frac{{\left( {v + {v_0}} \right)}}{{\left( {v – {v_s}} \right)}},\) \( \Rightarrow f = 2 \times \frac{{\left( {340 + 10} \right)}}{{\left( {340} \right)}} = 2.06\;{\rm{kHz}}\)

FAQs on Doppler Effect

Q.1. Is the observed frequency different than the natural frequency when the velocities are in direction perpendicular to the line joining the source and the observer?
Ans: No change is observed in the observed and the natural frequency when the velocities are in a perpendicular direction. Doppler effect only exists when the velocities are in the direction of the line joining the source and the observer.

Q.2. Is Doppler effect applicable for all waves?
Ans: Yes, Doppler effect is applicable for all waves, be it a longitudinal wave like a sound wave or a transverse wave like a light wave.

Q.3. Is doppler effect applicable for light waves?
Ans: Yes, doppler effect is applicable for light waves. Blue Shift or Red Shift phenomenon is observed in light waves due to doppler effect.

Q.4. What is doppler shift?
Ans: The shift in wavelength of the light towards red or blue in the visible light spectrum due to the motion of the stars is known as Doppler Shift.

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