# Doppler Effect: Definition, Equation & Example

You have likely noticed that the pitch of sound waves changes if it is generated by a moving source, whether approaching you or moving away from you.

For example, imagine standing on the sidewalk and hearing the sirens from an emergency vehicle approach and drive past. The frequency, or pitch of the siren as the vehicle approaches, is higher until it moves past you, at which point it becomes lower. The reason for this is something called the Doppler effect.

## What Is the Doppler Effect?

The Doppler effect, named for Austrian mathematician Christian Doppler, is a change in sound frequency (or the frequency of any wave, for that matter) caused because the source emitting the sound (or the observer) moves in the time between the emission of each successive wave front.

This results in an increase in the spacing of the wave peaks if it is moving away, or a decrease in the spacing of the wave peaks if a sound source is moving toward the observer.

Note that the speed of the sound in air does NOT change as a result of this motion. Only the wavelength, and hence the frequency, does. (Recall that wavelength *λ* , frequency *f* and wave speed *v* are related via *v = λf*.)

## Sound Source Approaching

Imagine a source emitting a sound of frequency _f_{source}* is moving toward a stationary observer with speed *v_{source}*. If the initial wavelength of the sound was *λ_{source}*, the wavelength detected by the observer should be the original wavelength *λ_{source}* minus how far the source moves during the time it takes to emit one full wavelength, or how far it moves in one period, or 1/*f_{source}_ seconds:

\(\lambda_{observer} = \lambda_{source} – \frac{v_{source}}{f_{source}}\)

Rewriting _λ_{source}* in terms of the speed of sound, *v_{sound}* and *f_{source}_ you get:

\(\lambda_{observer} = \frac{v_{sound}}{f_{source}} – \frac{v_{source}}{f_{source}}=\frac{v_{sound} – v_{source}}{f_{source}}\)

Using the fact that wave speed is the product of wavelength and frequency, you can determine what frequency the observer detects, _f_{observer}*, in terms of the speed of sound *v_{sound}_, the speed of the source, and the frequency emitted by the source.

\(f_{observer} = \frac{v_{sound}}{\lambda_{source}} = \frac{v_{sound}}{v_{sound} – v_{source}}f_{source}\)

This explains why sound seems to have higher pitch (higher frequency) when an object approaches you.

## Sound Source Receding

Imagine a source emitting a sound of frequency _f_{source}* is moving away from an observer with speed *v_{source}*. If the initial wavelength of the sound was *λ_{source}*, the wavelength detected by the observer should be the original wavelength *λ_{source}* plus how far the source moves during the time it takes to emit one full wavelength, or how far it moves in one period, or 1/*f_{source}_ seconds:

\(\lambda_{observer} = \lambda_{source} + \frac{v_{source}}{f_{source}}\)

Rewriting _λ_{source}* in terms of the speed of sound, *v_{sound}* and *f_{source}_ you get:

\(\lambda_{observer} = \frac{v_{sound}}{f_{source}} + \frac{v_{source}}{f_{source}} = \frac{v_{sound} + v_{source}}{f_{source}}\)

Using the fact that wave speed is the product of wavelength and frequency, you can determine what frequency the observer detects, _f_{observer}*, in terms of the speed of sound *v_{sound}_, the speed of the source, and the frequency emitted by the source.

\(f_{observer} = \frac{v_{sound}}{\lambda_{source}} = \frac{v_{sound}}{v_{sound} + v_{source}}f_{source}\)

This explains why sounds seem to have lower pitch (lower frequency) when a moving object is receding.

## Relative Motion

If both the source and the observer are moving, then the observed frequency depends on the relative velocity between the source and observer. The equation for observed frequency then becomes:

\(f_{observer} = \frac{v_{sound} ± v_{observer}}{v_{sound} ∓ v_{source}}f_{source}\)

The top signs being used for moving toward, and the bottoms signs being used for moving apart.

## Sonic Boom

As a high speed jet approaches the speed of sound, the sound waves in front of it start to "pile up" as their wave peaks become closer and closer together. This creates a very large amount of resistance as the plane attempts to reach and exceed the speed of sound.

Once the plane pushes through and surpasses the speed of sound, a shock wave is created and a very loud sonic boom results.

As the jet continues flying faster than the speed of sound, all sound associated with its flight lags behind it as it soars.

## Doppler Shift for Electromagnetic Waves

The Doppler shift for light waves works in much the same way. Approaching objects are said to demonstrate a blue shift since their light will be shifted toward the blue end of the em spectrum, and objects that are receding are said to demonstrate a red shift.

You can determine things such as the velocities of objects in space and even the expansion of the universe from this effect.

## Examples to Study

**Example 1:** A police car approaches you with its sirens blaring at a speed of 70 mph. How does the actual siren's frequency compare to the frequency you perceive? (Assume the speed of sound in air is 343 m/s)

First, convert 70 mph to m/s and get 31.3 m/s.

The frequency experienced by the observer is then:

\(f_{observer} = \frac{343\text{ m/s}}{343\text{ m/s} – 31.3\text{ m/s}}f_{source} = 1.1f_{source}\)

Hence you hear a frequency that is 1.1 times as great (or 10 percent higher) than the source frequency.

**Example 2:** 570 nm yellow light from an object in space is red shifted by 3 nm. How fast is this object receding?

Here you can use the same Doppler shift equations, but instead of _v_{sound}*, you would use *c_, the speed of light. Rewriting the observed wavelength equation for the light, you get:

\(\lambda_{observer} = \frac{c + v_{source}}{f_{source}}\)

Using the fact that _f_{source} = c/ λ_{source}*, and then solving for *v_{source}_, you get:

\(\begin{aligned}

&\lambda_{observer} = \frac{c + v_{source}}{c}\lambda_{source}\

&\implies v_{source} = \frac{\lambda_{observer} – \lambda_{source}}{\lambda_{source}}c\)

\(\end{aligned}\)

Finally, plugging in values, you get the answer:

\(v_{source} = \frac{3}{570}3\times 10^8\text{ m/s} = 1.58\times 10^6\text{ m/s}\)

Note that this is extremely fast (about 3.5 million miles per hour) and that even though the Doppler shift is called a "red" shift, this shifted light would still appear yellow to your eyes. The terms "red shifted" and "blue shifted" do not mean the light has become red or blue, but that it has simply shifted toward that end of the spectrum.

## Other Applications of the Doppler Effect

The Doppler effect is utilized in many different real world applications by scientists, doctors, the military and a whole host of other people. Not only that, but some animals have been known to make use of this effect to "see" by bouncing sound waves off of moving objects and listening to changes in pitch of the echo.

In astronomy, the Doppler effect is used to determine the rates of rotation of spiral galaxies and the speeds with which galaxies are receding.

The police make use of the Doppler effect with speed detecting radar guns. Meteorologists use it to track storms. Doppler echocardiograms used by doctors use sound waves to produce images of the heart and determine blood flow. The military even uses the Doppler effect to determine submarine speeds.

### Cite This Article

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TOWELL, GAYLE. "Doppler Effect: Definition, Equation & Example" *sciencing.com*, https://www.sciencing.com/doppler-effect-definition-equation-example-13722352/. 28 December 2020.

#### APA

TOWELL, GAYLE. (2020, December 28). Doppler Effect: Definition, Equation & Example. *sciencing.com*. Retrieved from https://www.sciencing.com/doppler-effect-definition-equation-example-13722352/

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TOWELL, GAYLE. Doppler Effect: Definition, Equation & Example last modified August 30, 2022. https://www.sciencing.com/doppler-effect-definition-equation-example-13722352/