Doppler Effect - Definition, Formula, Examples, Uses, FAQs

What is Doppler Effect Class 11 and Explain Doppler Effect?

Doppler effect definition: The Doppler Effect is the change in the frequency or wavelength of a wave about an observer who is moving relative to the wave source. It occurs when the source of the wave (such as sound, light, or other waves) and the observer are in motion relative to each other.

Whenever there is a relative motion between a source of sound and the listener, the apparent frequency/wavelength of sound heard by the listener is different from the actual frequency/wavelength of sound emitted by the source.

When the distance between the source and listener is increasing the apparent frequency decreases. It means the apparent frequency is less than the actual frequency of sound. The reverse of this process is also true.

Doppler Effect Formula

The general expression for apparent frequency

$$
f^{\prime}=f \frac{v \pm v_o}{v \mp v_s}
$$

where,

• $f^{\prime}$ is the observed frequency
• $f$ is the actual frequency of the source
• $v$ is the speed of sound in the medium
• $v_o$ is the speed of the observer
• $v_s$ is the speed of the source

Now, for different conditions, the value of apparent frequency will change.

There are some sign conventions for the velocities:

Along the direction Source to Listener is taken as positive and all velocities along the direction Listener to Source are taken as negative.

If the velocity of the medium is zero then the formula becomes

$$f^{\prime}=f \frac{v+v_{\text {listener }}}{v-v_{\text {source }}}$$

Also, read

• NCERT Solutions for All Subjects
• NCERT Notes for all subject
• NCERT Exemplar Solutions for All Subjects

Doppler Effect in Sound Waves Formula For Some Important Cases ($v$ is the speed of sound in medium)

(1) The source is moving towards the listener and the listener at rest then the formula becomes

$$
f^{\prime}=f\left(\frac{v}{v-v_{\text {source }}}\right)
$$

The sound waves are compacted as the source is moving towards the listener which increases the observed frequency in comparison with the source frequency.

(2) The source is moving away from the listener and the listener is at rest

$$
f^{\prime}=f\left(\frac{v}{v+v_{\text {source }}}\right)
$$

As the source is moving away from the listener the sound waves are spread thus decreasing the observed frequency relative to the original frequency.

(3) The source is at rest but the listener is moving away from the source $$
f^{\prime}=f\left(\frac{v-v_{\text {listener }}}{v}\right)
$$

$f^{\prime}$ decreases compared to the source frequency $f$

(4) The source is at rest but the listener is moving toward the source

$$
f^{\prime}=f\left(\frac{v+v_{\text {listener }}}{v}\right)
$$

As the listener is moving towards the source, observed frequency $f^{\prime}$ increases compared to the actual frequency.

(5) When the Source and listener are approaching each other

$$
f^{\prime}=f\left(\frac{v+v_{\text {listener }}}{v-v_{\text {source }}}\right)
$$

The observed frequency $f^{\prime}$ increases as compared to source frequency when both the source and listener are moving towards each other

(6) When the Source and listener move away from each other

$$
f^{\prime}=f\left(\frac{v-v_{\text {listener }}}{v+v_{\text {source }}}\right)
$$

The observed frequency $f^{\prime}$ decreases compared to the source frequency when the source and the listener are moving away from each other.

Note - Source and listener move perpendicular to the direction of sound wave propagation i.e., $f^{\prime}$ = $f$. It means there is no change in the frequency of the sound heard for the small displacement of the source and listener at a right angle to the direction of wave propagation but this is not true for large displacement. For a large displacement, the frequency decreases because the distance between the source of sound and the listener increases.

Doppler Effect in Light

In light of the observed frequency changes in connection with the relative motion of the source and listener. This is seen in two phenomena- red shift and blue shift.

Doppler effect in a light formula

$$
f^{\prime}=f \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}
$$

OR

$$
\lambda^{\prime}=\lambda \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}
$$

where,

• $f^{\prime}$ is the observed frequency
• $f$ is the source frequency
• $\lambda^{\prime}$ is the observed wavelength
• $\lambda$ is the source wavelength
• $v$ is the relative velocity between the source and observer
• $c$ is the speed of light in a vacuum

Red Shift

Consider the source is moving away from the observer. The observed wavelength increases while the source moves away shifting towards the red end of the light spectrum and frequency decreases. This phenomenon is called redshift.

$$
z=\frac{\lambda^{\prime}-\lambda}{\lambda}=\frac{v}{c}
$$
Using relativistic formula,

$$
z=\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}-1
$$

Blue Shift

When the source is moving towards the observer the observed frequency becomes shorter and shifts to the blue side of the spectrum.

$$
z=\frac{\lambda-\lambda^{\prime}}{\lambda}=-\frac{v}{c}
$$

Using relativistic formula,

$$
z=1-\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}
$$

where,

• $z$ is the redshift parameter
• $\lambda^{\prime}$ is the observed wavelength (longer, shifted towards red)
• $\lambda$ is the original wavelength of the source
• $v$ is the relative velocity of the source moving away from the observer
• $c$ is the speed of light in a vacuum

Application Of Doppler Effect

The Doppler effect is used in the following ways:

1. The Doppler effect (redshift and blueshift) is used to find the relative motion of celestial bodies.
2. It helps in identifying the speed, distance, and direction of planets and stars relative to Earth.
3. The Doppler effect is used in Doppler ultrasound to measure blood flow and heart function.
4. The Doppler effect is used in LIDAR systems
5. The Doppler effects are also used to study seismic waves.

Limitations Of the Doppler Effect

1. The Doppler effect only happens when there is a relative motion between the source and the observer.
2. The Doppler effect in sound waves depends on the medium.
3. The Doppler effect formula is inaccurate at high relative velocities which is close to the speed of light.

Recommended Topic Video

Solved Examples Based on the Doppler Effect

Example 1: An observer is moving with half the speed of light towards a stationary microwave source emitting waves at a frequency of 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light=3×10⁸ ms⁻¹)

1) 10.1 GHz

2) 12.1 GHz

3) 17.3 GHz

4) 15.3 GHz

Solution:

Doppler Effect

When a source of sound and a listener are in motion relative to each other, the frequency of the sound heard by the listener is not the same as the source frequency.

Doppler effect in light

$v^{\prime}=v \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}=17.3 \mathrm{GHz}$

Hence, the answer is the option (3).

Example 2: An observer moves towards a stationary source of sound, with a velocity one­ fifth of the velocity of sound. What is the percentage( in % ) increase in the apparent frequency?

1) 20

2) 5

3) 0

4) 0.5

Solution:

By Doppler's effect

$\begin{gathered}v^{\prime}=\nu\left(\frac{v_s+v_0}{v_s}\right)=\nu+\frac{\nu}{5} \\ \qquad v^{\prime}=\frac{6 \nu}{5} \\ \text { Fractional increase }=\frac{v^{\prime}-v}{v}=\frac{6}{5}-1=\frac{1}{5} \\ \text { Percentage increase }=\frac{1}{5} \times 100=20 \%\end{gathered}$

Hence, the answer is the option (1).

Example 3: Two sources of sounds S1 and S2 produce sound waves of the same frequency 660 Hz. A listener is moving from source S1 towards S2 with a constant speed of u m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then, u (in m/s) equals:

1) 2.5

2) 15.0

3) 5.5

4) 10

Solution:

Doppler Effect

When a source of the sound and a listener are in motion relative to each other, the frequency of the sound heard by the listener is not the same as the source frequency.

$\begin{gathered}f_0=660 \mathrm{H} 2 \\ f_1=\text { apparent frequency received from } \mathrm{S} 1=\frac{C-u}{C} f_0 \\ f_2=\text { apparent frequency received from } \mathrm{S} 2=\frac{C+u}{C} f_0 \\ f_2-f_1=\left[\frac{C+u}{C}-\frac{C-u}{C}\right] f_0 \\ \text { beat frequency }=\frac{2 u}{C} f_0=10 \\ \frac{2 u \times 660}{330}=10 \\ u=\frac{10 \times 330}{2 \times 660}=2.5 \mathrm{~m} / \mathrm{s}\end{gathered}$

Hence, the answer is the option (1).

Example 4: A source of sound emits sound waves at frequency f0. It is moving towards an observer with fixed speed vs(vs<v, where v is the speed of sound in air) If the observer were to move towards the source with speed v0, one of the following two graphs (A and B) will give the correct variation of the frequency f heard by the observer as v0 is changed.

The variation of f with v0 is given correctly by :

1. Graph A with slope $=f_0\left(\nu-\nu_s\right)$
2. Graph A with slope $=f_0\left(\nu+\nu_s\right)$
3. Graph B with slope $=f_0\left(\nu-\nu_s\right)$
4. Graph B with slope $=f_0\left(\nu+\nu_s\right)$

Solution:

Frequency of sound when source and observer are moving toward each other

$
\begin{aligned}
& \qquad \nu^{\prime}=\nu_0 \cdot \frac{C+V_0}{C-V_s} \\
& \text { wherein } \\
& C=\text { Speed of sound } \\
& V_0=\text { Speed of observer } \\
& V_s=\text { Speed of source } \\
& \nu_0=\text { Original Frequency } \\
& \nu^{\prime}=\text { apparent frequency }
\end{aligned}
$

Graph A with slope $=f_0\left(\nu-V_s\right)$

Hence, the answer is the option (1).

Example 5: A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed $\nu$ ms^-1 The velocity of sound in air is 300 ms^-1. If the person can hear frequencies up to a maximum of 10000 Hz, the maximum value of ν up to which he can hear the whistle is :

1) 30 ms−1
2) 152 ms−1
3) 15/2 ms−1
4) 15 ms−1

Solution:

Frequency of sound when the observer is stationary and the source is moving towards the observer -

$
\nu^{\prime}=\nu_0 \cdot \frac{C}{C-V_s}
$

where
$C=$ the speed of sound
$V_s=$ speed of source
$\nu_0=$ original frequency
$\nu^{\prime}=$ apparent frequency

$
\frac{v^{\prime}}{v}=\frac{V_s}{V_s-v}
$

Where $V_s$ is the velocity of sound in air.

$\begin{gathered}\frac{10000}{9500}=\frac{300}{300-v} \\ (300-v)=285 \\ v=15 \mathrm{~m} / \mathrm{s}\end{gathered}$

Hence the answer is the option (4).