**The Doppler Effect**

Austrian physicist Christian Doppler discovered in his study of sound waves that the frequency and pitch of sound can change due to the motion of the either the source or the receiver of the sound. This is known as the **Doppler effect.**

**Frequency** is the amount of sound waves produced per second. **Pitch **is the lowness or highness of sound. Pitch depends on the frequency of the sound. Faster sound waves denotes greater frequency and higher pitch.

**If the source and the receiver of the sound are not moving**, the frequency of sound as it was produced is the same as how it is heard by the receiver. The wavelength is also the same from the source to the receiver. Thus, the Doppler effect does not happen if the source and the receiver are at rest.

**If the source is moving towards the receiver**, the frequency of the sound heard by the receiver is higher than the frequency of the sound source. The wavelength decreases, hence, more crests are encountered by the receiver, making the sound heard louder by the receiver. This can be expressed in the equation as,

\(f=f_s \begin{pmatrix} \frac {v}{v-v_s} \end{pmatrix}\)

where

\(f=\text{frequency of the sound as heard by the receiver}\\ f_s=\text{frequency of the sound waves}\\ v=\text{velocity of the sound waves}\\ v_s=\text{velocity of the source}\)

**If the source is moving away from the receiver**, the frequency of the sound as heard by the receiver is lesser. It is mathematically expressed as

\(f=f_s \begin{pmatrix} \frac {v}{v+v_s} \end{pmatrix}\)

**When the receiver is moving towards the source**, the velocity of the receiver is positive and will hear a higher frequency of sound. In equation form,

\(f=f_s \begin{pmatrix} \frac {v+v_r}{v} \end{pmatrix}\)

\(v_r\) is the velocity of the receiver.

**When the receiver is moving away from the source**, the velocity of the receiver is negative, thus the frequency of the sound heard is lesser than the frequency of the sound waves.

\(f=f_s \begin{pmatrix} \frac {v-v_r}{v} \end{pmatrix}\)

**Example 1.**

Eric is walking at the lobby when he passes by the music class. The teacher is playing the piano which emits a sound with frequency of \(800 \;Hz\). When he walks at a speed of \(1.5\;m/s\), what is the frequency of the piano as he approaches the class? What is the frequency of the piano’s sound as heard by Eric after passing the music room? The recorded temperature that time is \(25\;^\circ C\).

Given:

\(f_p=800\;Hz\\ v_E=1.5\;m/s\\ T=25\;^\circ C\)

Solution 1:

Solve first for the speed of sound in air at \(25\;^\circ C\).

\(v=331\;m/s+\frac{0.6\;m/s}{^\circ C}(25^\circ C)=346\;m/s\)

Solution 2:

\(f=f_p \begin{pmatrix} \frac {v+v_E}{v} \end{pmatrix}\\ \quad=800 \;Hz \begin{pmatrix} \frac {346\;m/s+1.5\;m/s}{346\;m/s} \end{pmatrix}\\ \quad=800\;Hz \begin{pmatrix} \frac{347.5\;m/s}{346\;m/s} \end{pmatrix}\\ \quad=800\;Hz\;(1.004)\\ \quad=803.2\;Hz\)

The frequency of the sound of the piano heard by Eric as he approaches the music room is \(803.2\;Hz\).

Solution 3:

\(f=f_p \begin{pmatrix} \frac {v-v_E}{v} \end{pmatrix}\\ \quad=800 \;Hz \begin{pmatrix} \frac {346\;m/s-1.5\;m/s}{346\;m/s} \end{pmatrix}\\ \quad=800\;Hz \begin{pmatrix} \frac{344.5\;m/s}{346\;m/s} \end{pmatrix}\\ \quad=800\;Hz(0.996)\\ \quad=796.8\;Hz\)

The frequency of the sound of the piano heard by Eric as he approaches the music room is \(796.8\;Hz\).

**Example 2.**

Mark is standing at the roadside when he saw a fire truck approaching at \(28\;m/s \). If the frequency of the truck’s siren is \(1, 350 \;Hz\), what is the frequency of the siren heard by Mark as the truck is approaching? How much frequency of sound Mark heard after the truck left the point where Mark is standing? The speed of sound in air is \(342\;m/s\).

Given:

\(f_s=1,350\;Hz\\ v_{ft}=28\;m/s\\ v=342\;m/s\)

Solution 1:

\(f=f_s \begin{pmatrix} \frac {v}{v-v_s} \end{pmatrix}\\ \quad=1,350\;Hz \begin{pmatrix} \frac {342\;m/s}{342\;m/s-28\;m/s} \end{pmatrix}\\ \quad=1,350\;Hz \begin{pmatrix} \frac {342\;m/s}{314\;m/s} \end{pmatrix}\\ \quad=1,350\;Hz(1.09)\\ \quad=1,471.50\;Hz\)

The frequency of the siren heard by Mark as the truck approaches is \(1, 471.50\;Hz\).

Solution 2.

\(f=f_s \begin{pmatrix} \frac {v}{v+v_s} \end{pmatrix}\\ \quad=1,350\;Hz \begin{pmatrix} \frac {342\;m/s}{342\;m/s+28\;m/s} \end{pmatrix}\\ \quad=1,350\;Hz \begin{pmatrix} \frac {342\;m/s}{370\;m/s} \end{pmatrix}\\ \quad=1,350\;Hz(0.92)\\ \quad=1, 242\;Hz\)

The frequency of the siren heard by Mark after the truck pass him is \(1, 242\;Hz\).