You all must have heard the whistle of a moving train. What do you feel when the train approaches you? The pitch of the whistle seems to rise. But when the engine passes by, the pitch appears to decrease. Similarly, while standing near a highway you may have also noticed that a loaded truck approaching you makes a relatively high-pitched sound ‘aaaaaaaa’. As trucks passes you, the pitch drops abruptly ‘aaaaaaeeeeiiioooooo’ and stays low as the truck recedes. The apparent change of frequency due to the relative motion between the source and the observer (or the listener) is known as the Doppler Effect.
In general, when the source approaches the listener or the listener approaches the source, or both approach each other, the apparent frequency is higher than the actual frequency of the sound produced by the source. Similarly when the source moves away from the listener, or when the listener moves away from the source, or when both move away from each other, the apparent frequency is lower than the actual frequency of the sound produced by the source.
Do you know that Doppler shift in ultra-sound reflected from moving body tissues allows measurement of blood flow? It is commonly used by obstetricians to detect foetal heart-beat.
The electromagnetic waves, including light, are also subject to the Doppler effect. In air navigation, radar works by measuring the Doppler shift of high frequency radio waves reflected from moving aeroplanes. The Doppler shift of star-light allows us to study steller motion. When we examine light from stars in a spectrograph, we observer several spectral lines. These lines are slightly shifted as compared to the corresponding lines from the same elements on the earth. This shifts is generally towards the red-end and is attributed to steller motion.
To study the Doppler effect for sound waves, we have to consider the following situation.
(i) Whether the source is in motion, or the observer is in motion, or both are in motion.
(ii) Whether the motion is along the line joining the source and the observer, or inclined (at an angle) to it.
(iii) Whether the direction of motion of the medium is along or opposite to the direction of propagation of sound.
(iv) Whether the speed of source is greater or smaller than the speed of sound produced by it.
We will now consider some of these possibilities.
Source in Motion and Observer Stationary:
Let us suppose that a source S is producing sound of frequency n, and wavelength l. The waves emitted by the source spread out as spherical wavefronts of sound. When the velocity of source is less than the velocity of sound, wavefronts lie inside one another. The distance between successive wavefronts is minimum along the direction of motion and maximum in a direction opposite to it.
Representing the same situation in terms of waves, as shown in figure (a), we find that if v is the speed of sound produced, n waves occupy a length v in one second, if the source is stationary. After one second, when the source has moved a distance us towards the listener, the same number of waves get crowded a length (v – us) as shown in figure (b).
The reduced wavelength l’, then becomes
\({\lambda }'\frac{v-{{u}_{s}}}{\nu }\)
The apparent frequency of sound (heard by the listener) is then
\({\nu }'=\frac{u}{{{\lambda }'}}=\left( \frac{v}{v-{{u}_{s}}} \right)\nu \)
Source Stationary and observer in motion:
If the observer is at rest, the length of the block of waves passing him per second is v and contains n waves. However, when the observer moves with speed uu, he will be at O’ after one second and find that only a block of waves with length (v – uu) passes him in one second. For him the apparent frequency is then
\({v}'=\frac{v-{{u}_{0}}}{\lambda }=v\frac{v-{{u}_{0}}}{v}\) … (1)
If the listener moves towards the source, u0 is negative, and the apparent frequency is given by
\({v}'=v\frac{v+{{u}_{0}}}{v}\) … (2)
Source and Observer both in Motion:
When both source and observer are in motion (and approach each other), we have to combine the result obtained in equations (1) and (2). The source in motion causes a change in wavelength. The listener in motion results in a change of number of waves received. In such a case, apparent frequency v’ is given by
n’ = \(\frac{\text{Length of block of waves received}}{\text{Reduced wavelength}}\)
\({\nu }'=\left( \frac{v+{{u}_{0}}}{v-{{u}_{s}}} \right)\nu \)
You may now ask: Is there any difference in the apparent frequency when the source approaches the listener or the listener approaches the source with the same velocity? (equation \(\left( \frac{v+{{u}_{0}}}{v-{{u}_{s}}} \right)\) tells us that the apparent frequency will be different in these cases.
For electromagnetic waves, equation has to be modified. For sound, u0 and us are measured relative to the medium. This is because the medium determines the wave speed. However, e.m. waves do not require a medium for propagation so that their speed relative to source or the observer is always the same. For these waves we have to consider only the relative motion of the source and the observer. If us is the speed of source relative to observer, and \({{u}_{s}}\ll c\) (c – speed of electromagnetic wave) we can rewrite equation
\({v}'=\left( \frac{c+{{u}_{0}}}{c-{{u}_{s}}} \right)v\)
Þ
Using binomial expansion and retaining only first order terms in (us / v), we get
\({\nu }'=v\left( 1+\frac{{{u}_{s}}}{c} \right)\)
In air navigation, we take us to be twice the approach velocity of the aeroplane. This is because the radar detects e.m. waves sent by it and reflected back by the aeroplane.
1. A source of sound moves away with velocity of sound from a stationary observer. Then the frequency of sound heard by the observer
Solution:
Is halved
2. In Doppler effect, when a source moves towards a stationary observer, the apparent increase in frequency is due to
Solution:
Decrease in wavelength of sound received by observer
3. Doppler shift in frequency does not depend upon
Solution:
Distance between source and observer
1. An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. The percentage increase in the apparent frequency is
Solution:
\(f={{f}_{0}}\left[ \frac{{{v}_{s}}+{{v}_{0}}}{{{v}_{s}}} \right]={{f}_{0}}\left[ \frac{v+\frac{v}{5}}{v} \right]=\frac{6}{5}{{f}_{0}}\)
Percentage increase \(=\left[ \frac{\frac{6}{5}{{f}_{0}}-{{f}_{0}}}{{{f}_{0}}} \right]\times 100=20%\)
= 20%
2. A whistling engine is approaching a stationary observer with a velocity of 110m/s. The velocity of sound is 330m/s. The ratio of frequencies as heard by the observer as the engine approaches and receedes is
Solution:
\(n_{app}^{'}=\left( \frac{V}{V-{{V}_{s}}} \right)n\,;\ \ \ n_{rec}^{'}=\left( \frac{V}{V+{{V}_{s}}} \right)n\)
= 2:1
1. A siren is fitted on a car going towards a vertical wall at a speed of 36 km/h. A person standing on the ground, behind the car, listens to the siren sound coming directly from the source as well as that coming after reflection from the wall. Calculate the apparent frequency of the wave (a) coming directly from the siren to the person.
Solution:
The speed of the car is 36 km/h = 10 m/s.
(a) Here the observer is at rest with respect to the medium and the source is going away from the observer. The apparent frequency heard by the observer is, therefore,
\({v}'=\frac{v}{v+{{u}_{s}}}v\)
\(=\frac{340}{340+10}\times 500\ \text{Hz}=486\ \text{Hz}.\)
2. A siren is fitted on a car going towards a vertical wall at a speed of 36 km/h. A person standing on the ground, behind the car, listens to the siren sound coming directly from the source as well as that coming after reflection from the wall. Calculate the apparent frequency of the wave (b) coming after reflection. Take the speed of sound to be 340 m/s.
Solution:
(b) The frequency received by the wall is
\({v}''=\frac{v}{v-{{u}_{s}}}v=\frac{340}{340-10}\times 500\ \text{Hz}=515\ \text{Hz}.\)
The wall reflects this sound without changing the frequency. Thus, the frequency of the reflected wave as heard by the ground observer is 515 Hz.
1. A train approaching a railway crossing at a speed of 120 km/h sounds a short whistle at frequency 640 Hz when it is 300 m away from the crossing. The speed of sound in air is 340 m/s. What will be the frequency heard by a person standing on a road perpendicular to the track through the crossing at a distance of 400 m from the crossing?
Solution:
The observer A is at rest with respect to the air and the source is traveling at a velocity of 120 km/h i.e., \(\frac{100}{3}\text{m/s}.\) As is clear from the figure, the person receives the sound of the whistle in a direction BA making an angle q with the track where cos q = 300/500 = 3/5. The component of the velocity of the source (i.e., of the train) along the direction AB is \(\frac{100}{3}\times \frac{3}{5}\text{m/s}=20\ \text{m/s}.\) As the source is approaching the person with this component, the frequency heard by the observer is
\({v}'=\frac{v}{v-u\cos \theta }v=\frac{340}{340-20}\times 640\ \text{Hz}=680\ \text{Hz}.\)
2. A stationary source emits a sound towards a wall moving towards it with a velocity u. Speed of sound in air = v. Find the fractional change in wavelength of the sound sent and the reflected sound.
Solution:
Let the frequency of the sound sent by the source be fo.
Frequency of sound as observed by the wall is
\(\therefore \,\,\,\,\,\,\,f'=\left( \frac{v+u}{v} \right)\,{{f}_{o}}\)
Frequency of the sound reflected by the wall is equal that received by it.
Now, the frequency f¢¢ received by the source is
\(\begin{align} & \,\,\,\,\,\,\,f''=\left( \frac{v}{v-u} \right)\,f' \\ & \,\,\,\,\,\,\,f''=\left( \frac{v}{v-u}\times \frac{v+u}{v} \right){{f}_{o}} \\ & \,\,\,\,\,\,\,\,f''=\left( \frac{v+u}{v-u} \right)\,\,{{f}_{o}} \\ \end{align}\)
\(\begin{align} & \text{Initial wavelength}\,\,\,\,\,{{\lambda }_{\text{i}}}=\frac{v}{{{f}_{o}}} \\ & \text{Final wavelength }\,\,{{\lambda }_{\text{f}}}=\frac{v}{f\,''}=\frac{v}{{{f}_{o}}}\left( \frac{v-u}{v+u} \right) \\ \end{align}\)
Fractional change in wavelength = \(\frac{\Delta \lambda }{{{\lambda }_{i}}}=\frac{2u}{v+u}\)