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

The Doppler Effect is a fascinating phenomenon observed when the source of a sound or light wave moves relative to an observer, leading to a change in the perceived frequency of the wave. Named after the Austrian physicist Christian Doppler, who first proposed it in 1842, this effect is commonly experienced in everyday life. For instance, when an ambulance with a blaring siren speeds past you, the pitch of the siren appears higher as it approaches and lower as it moves away. This shift in frequency isn't just limited to sound; it's also critical in astronomy for measuring the speed and direction of distant stars and galaxies. The Doppler Effect is also used in radar technology, medical imaging, and even in determining the velocity of blood flow in arteries, making it a crucial concept with wide-ranging real-life applications. In this article, we will discuss the concept of the Doppler Effect with solved examples.

What is the Doppler Effect?

The Doppler Effect is the change in the frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. It occurs when the source of the wave (such as sound, light, or other types of 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.

General expression for apparent frequency $f_{a p p}=\frac{\left[\left(v+v_m\right)-v_o\right] f}{\left[\left(v+v_m\right)-v_s\right]}$

Now, for different conditions, the value of apparent frequency will change. Here f = Actual frequency; v₀= Velocity of observer; v_S = Velocity of source, v_M = Velocity of medium and v = Velocity of sound wave

There are some sign conventions for the velocities - along the direction Source to Listener are 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 $f_{a p p}=\left(\frac{v-v_o}{v-v_s}\right) f$

Now we will discuss some important cases and based on that the formulas

(1) Source is moving towards the listener and the listener at rest $f_{a p p}=\frac{v}{v-v_s} \cdot f$
(2) Source is moving away from the listener and the listener is at rest $f_{a p p}=\frac{v}{v+v_s} \cdot f$
(3) Source is at rest but the listener is moving away from the source $f_{a p p}=\frac{v-v_L}{v} f$
(4) Source is at rest but the listener is moving towards the source $f_{\text {app }}=\frac{v+v_L}{v} f$
(5) When the Source and listener are approaching each other $f_{\text {app }}=\left(\frac{v+v_L}{v-v_S}\right) f$
(6) When the Source and listener move away from each other $f_{a p p}=\left(\frac{v-v_L}{v+v_s}\right) f$

Note - Source and listener move perpendicular to the direction of wave propagation i.e., f_app = 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.

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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

$\begin{aligned} & v^{\prime}=\sqrt{\frac{1+v / c}{1-v / c}} \Rightarrow \sqrt{\frac{1+1 / 2}{1-1 / 2}} \\ & v^{\prime}=17.3 \mathrm{GHz}\end{aligned}$

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

$\begin{aligned} & \frac{v^{\prime}}{v}=\frac{\nu_s+\nu_0}{\nu_s} \quad\left(\text { where } v_s \text { is the velocity of sound) }\right. \\ & =\frac{\nu+(\nu / 5)}{\nu}=\frac{6}{5} \\ & \therefore \quad \text { Fractional increase } \\ & \qquad \frac{v^{\prime}-v}{v}=\left(\frac{6}{5}-1\right)=\frac{1}{5} \\ & \therefore \quad \text { Percentage increase }=\frac{100}{5}=20 \%\end{aligned}$

Hence, the answer is the option (1).

Example 3: Two sources of sounds $S_1$ and $S_2$ produce sound waves of the same frequency 660 Hz. A listener is moving from source $S_1$ towards $S_2$ with a constant speed 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.

$f_0=660 H_2$ $f_0=660 H_2$

$\begin{aligned} & f_1=\text { apparent frequency received from } S_1=\frac{C-u}{C} f_0 \\ & f_2=\text { apparent frequency received from } S_2=\frac{C+u}{C} f_0 \\ & \qquad f_2-f_1=\left[\frac{C+u}{C}-\frac{C-u}{C}\right] f_0 \\ & \text { beat frequency } \\ & =\frac{2 u}{C} f_0=10 \\ & \frac{2 u \times 660}{330}=10 \\ & u=\frac{10}{4}=2.5 \mathrm{~m} / \mathrm{s}\end{aligned}$

Hence, the answer is the option (1).

Example 4: A source of sound emits sound waves at frequency $f_0$. It is moving towards an observer with fixed speed $v_s\left(v_s<v\right.$, where $v$ is the speed of sound in air) If the observer were to move towards the source with speed $v_0$, one of the following two graphs (A and B) will give the correct variation of the frequency f heard by the observer as $v_0$ is changed.

The variation of f with $v_0$ is given correctly by :

1) graph A with slope $=\frac{f_0}{\left(\nu-\nu_s\right)}$
2) graph A with slope $=\frac{f_0}{\left(\nu+\nu_s\right)}$
3) graph $B$ with slope $=\frac{f_0}{\left(\nu-\nu_s\right)}$
4) graph $B$ with slope $=\frac{f_0}{\left(\nu+\nu_s\right)}$

Solution:

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

$
\begin{aligned}
& \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 } \\
& \text { Graph A with slope }=\frac{f_0}{\left(v-v_s\right)}
\end{aligned}
$

Graph A with slope

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 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 $\nu$ upto which he can hear the whistle is :

1) $30 \mathrm{~ms}^{-1}$
2) $15 \sqrt{2} \mathrm{~ms}^{-1}$
3) $15 / \sqrt{2} \mathrm{~ms}^{-1}$
4) $15 \mathrm{~ms}^{-1}$

Solution:

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

$
\begin{aligned}
& \nu^{\prime}=\nu_0 \cdot \frac{C}{C-V_s} \\
& \text { where C }=\text { the speed of sound } \\
& V_s=\text { speed of source } \\
& \nu_0=\text { original frequency } \\
& \nu^{\prime}=\text { apparent frequency } \\
& \frac{v^{\prime}}{v}=\frac{v_s}{v_s-v}
\end{aligned}
$

Where $v_s$ is the velocity of sound in air.
$
\begin{aligned}
& \frac{10000}{9500}=\frac{300}{300-v} \\
& \Rightarrow(300-v)=285 \Rightarrow v=15 \mathrm{~m} / \mathrm{s}
\end{aligned}
$

Hence the answer is the option (4).

Summary

The Doppler Effect describes the change in frequency of waves (sound or light) due to the relative motion between the source and the observer. This phenomenon explains why sounds appear higher in pitch when approaching and lower when receding. The concept is crucial for various applications, from medical imaging and radar technology to understanding cosmic phenomena. Through different examples, the article illustrates how the Doppler Effect's principles apply in real-world scenarios, emphasizing its significance in both everyday life and scientific studies.