# Doppler effect: description, formulas, cases, examples

We explain the Doppler effect examples with description, formulas, cases and definition. The **Doppler effect** is a physical phenomenon that occurs when the receiver and the source of waves have relative movement, causing a change in the frequency of the receiver with respect to the frequency of the source. example of doppler effect

Its name comes from the Austrian physicist Christian Doppler (1803-1853), who described and explained this phenomenon in 1842, while presenting a work on the color of double stars, at a congress of natural sciences in Prague, present-day Czech Republic.

**Where is the Doppler effect present?**

The Doppler effect occurs on all types of waves, from light to sound, as long as the source and receiver are moving relative to each other.Â And it is much more remarkable when the relative speed between the source and receiver is comparable to the speed of propagation of the wave.

Suppose a harmonic wave, which is an oscillation moving through space.Â The oscillation is repeated at regular time intervals, this time isÂ *the period*Â and its inverseÂ *the frequency* , that is, the number of oscillations per unit of time.

When the distance between the source of the harmonic wave and the receiver remains fixed, the receiver perceives the same frequency of the source, that is, it registers the same number of pulses per unit time as the source. example of doppler effect

However, when the receiver approaches the source with a fixed speed, then the pulses arrive more frequently.Â And the opposite occurs when the receiver moves away at a fixed speed from the source: the wave pulses are perceived with a lower frequency.

**Description of the Doppler effect**

To understand why this phenomenon occurs we will use an analogy: two people playing throwing balls. The pitcher rolls them in a straight line across the ground toward his partner, who picks them up. example of doppler effect

If the person throwing sends a ball every second, the catcher, if he stays fixed, will catch a ball every second.Â All good so far, as it is expected.

**The receiver in motion**

Now suppose the person catching the balls is on a skateboard and decides to approach the pitcher with constant speed.Â In this case, since you are going to meet the balls, you will have less than a second between one ball and the next.

Therefore, it seems to the receiver that more than one ball is reaching him per second, that is, the frequency with which they reach his hand increased. example of doppler effect

The opposite would happen if the receiving person decided to move away from the emitter, that is, the arrival time of the balls would increase with the consequent decrease in the frequency with which the balls arrive.

**Formulas example of doppler effect**

The change in frequency described in the previous section can be obtained from the following formula:

Here:

-fÂ _{o}Â is the frequency of the source.

-f is the apparent frequency at the receiver.

-v is the speed (v> 0) of propagation of the wave in the medium.

-vÂ _{r}Â is the velocity of the receiver relative to the medium and

-vÂ _{s}Â Â is the velocity of the source relative to the medium.

Note that vÂ _{r}Â Â is positive if the receiver approaches the source and negative otherwise.Â On the other hand, vÂ _{s}Â Â is positive if the source moves away from the receiver and negative when it approaches.

Ultimately, if the source and the observer get closer, the frequency increases and if they move away it decreases. The opposite occurs with the apparent wavelength at the receiver (see exercise 1). example of doppler effect

**Cases in which the Doppler effect exists**

**Source and receiver speed much slower than the wave**

It often happens that the speed of the wave is vastly greater than the speed with which the source is moving or the speed of movement of the receiver.

In this case, the formula can be approximated in such a way that it is written as a function of the relative speed of the receiver (observer) with respect to the source (s).

In this case the formula would be like this:

f = [1 + (VÂ rsÂ / v)] â‹…fÂ _{or}

Where VÂ _{rs}Â = vÂ _{rÂ }Â – vÂ _{s}Â .

When VÂ _{rs}Â Â is positive (they approach), the frequency f is greater than fÂ _{o}Â , while when it is negative (they move away), f is less than fÂ _{o} . example of doppler effect

**Relative velocity at an angle to the relative position**

The above formula applies only to the case that the source approaches (or moves away) directly from the observer.

In case the source moves along a transversal path, it is necessary to take into account the angle Î¸ formed by the relative velocity of the receiver -with respect to the source- with the direction of the vector that goes from the observer to the source.

In this case we must apply:

f = [1 + (VÂ _{rs}Â Â â‹… Cos (Î¸) / v)] â‹… fÂ _{o}

Again, VÂ _{rs}Â Â is assigned a positive sign if the receiver and source are approaching, and aÂ Â negativeÂ sign if theÂ opposite is the case.

**Examples of Doppler effect**

An everyday example is the siren of an ambulance or police car.Â When it approaches us it is perceived more acute and when it moves away it is more serious, particularly the difference is heard at the moment of maximum approach.

Another situation that is explained by the Doppler effect is the shift of the spectral lines of the stars towards blue or red, if they are approaching towards us or if they are moving away.Â This cannot be seen with the naked eye, but with an instrument called aÂ *spectrometer* . example of doppler effect

**Applications example of doppler effect**

The Doppler effect has many practical applications, some are listed below:

**Radars**

Radars measure the distance and speed at which the objects detected by it move and are based precisely on the Doppler effect.

The radar emits a wave towards the object to be detected, then that wave is reflected back. The time it takes for a pulse to go back and forth is used to determine how far away the object is. And the change in frequency in the reflected signal allows you to know if the object in question is moving away from or approaching the radar and how fast. example of doppler effect

Because the radar wave goes back and forth, a double Doppler effect occurs.Â In this case, the formula for determining the speed of the object with respect to the radar is:

VÂ _{o / r}Â = Â½ c â‹… (Î”f / fÂ _{o}Â )

Where:

-VÂ _{o / r}Â is the speed of the object with respect to the radar.

-c the speed of the wave emitted and then reflected.

-fÂ _{or}Â the frequency of emission on the radar.

-Î”f the frequency shift, that is f – fÂ _{o}Â .

**Astronomy example of doppler effect**

Thanks to the Doppler effect, it has been possible to determine that the universe is expanding, since the light spectrum emitted by distant galaxies is shifted towards the red (a decrease in frequency).

On the other hand, it is also known that the receding speed increases as the observed galaxies are more distant.

The opposite case occurs with some galaxies of the local group, that is, the neighbors of our Milky Way.

For example, our closest neighbor, the Andromeda Galaxy, has a blue shift (that is, an increase in frequency) which indicates that it is approaching us. example of doppler effect

**Doppler ultrasound**

It is a variant of the traditional ecosonogram, in which, taking advantage of the Doppler effect, the speed of blood flow in veins and arteries is measured.

**Training example of doppler effect**

**Exercise 1Â **

The siren of an ambulance has a frequency of 300 Hz. Knowing that the speed of sound in air is 340 m / s, determine the wavelength of the sound in the following cases:

a) When the ambulance is at rest.

b) If it approaches 108 km / h

c) When moving away at the same speed.

**Solution to example of doppler effect**

There is no Doppler effect because both the emitter and the source are at rest.

To determine the wavelength of sound, the relationship between the frequency of the source f, the wavelength Î» of the source and the speed of sound v is used:

v = fÂ _{or}Â â‹…Î».

From there it follows that:

Î» = v / fÂ _{o}Â .

Therefore the wavelength is:

Î» = (340 m / s) / (300 1 / s) = 1.13 m.

**Solution b example of doppler effect**

The receiver is considered at rest, that is, vÂ _{r}Â Â = 0. The emitter is the siren that moves with the speed of the ambulance:

vÂ _{s}Â Â = (108 / 3.6) m / s = 30 m / s.

The apparent frequency f is given by the relation:

f = fÂ _{or}Â â‹… [(v + vÂ _{r}Â ) / (v + vÂ _{s}Â )]

Applying this formula we obtain:

f = 300 Hz â‹… [(340 + 0) / (340 – 30)] = 329 Hz.

The wavelength at the receiver will be:

Î»Â _{r}Â = v / f = (340 m / s) / (329 1 / s) = 1.03 m.

**Solution c**

It is solved in a similar way:

f = 300 Hz â‹… (340 + 0) / (340 + 30) = 276 Hz.

The wavelength at the receiver will be:

Î»Â _{r}Â = v / f = (340 m / s) / (276 1 / s) = 1.23 m.

It is concluded that the wave fronts have a separation of 1.03 m when the siren approaches and 1.23 m when it moves away.

**Exercise 2**

A characteristic line of the hydrogen emission spectrum is at 656 nm, but when observing a galaxy it is seen that that same line is displaced and marks 660 nm, that is, it has a redshift of 4 nm.

Since there is an increase in wavelength, we know that the galaxy is moving away.Â What is its velocity?

**Solution example of doppler effect**

The quotient between the displacement of the wavelength and the wavelength at rest is equal to the quotient between the speed of the galaxy and the speed of light (300,000 km / s).Â Then:

4/656 = 0.006

Therefore the galaxy is moving away at 0.006 times the speed of light, that is, at 1800 km / s.