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Class 12 Physics Chapter 9: Acoustic Phenomena Notes (NEB New Syllabus) | Sound Waves, Intensity, Loudness, Doppler Effect, Numerical Problems & Solutions | Acoustic Phenomena PDF
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1. Syllabus

Unit: Waves and Optics

Chapter 9: Acoustic Phenomena (5 hours)

9.1 Sound waves: pressure amplitude
9.2 Characteristics of sound: intensity; loudness, quality and pitch
9.3 Doppler’s effect

These notes follow the NEB Class 12 Physics syllabus (+2 Physics, Nepal). Practise the multiple choice questions first, then work through the numericals, and finish with the PDF notes for quick revision.

2. Multiple Choice Questions

Question 1. The intensity of a sound wave is directly proportional to which of the following?

  • a) Amplitude
  • b) Square of amplitude
  • c) Frequency
  • d) Wavelength

Correct Answer: b) Square of amplitude

Reason: Intensity is the average energy transported per unit time per unit area:

\[ I=\tfrac12\rho v\omega^2A^2 \]

where \(\rho\) is the density of the medium, \(v\) the wave speed, \(\omega\) the angular frequency and \(A\) the amplitude. Hence \(I\propto A^2\).

Question 2. The pitch of a sound depends primarily on:

  • a) Frequency
  • b) Intensity
  • c) Loudness
  • d) Quality

Correct Answer: a) Frequency

Reason: Pitch is the sensation of how high or low a sound is. It is determined mainly by the frequency of the sound wave: the higher the frequency, the higher the pitch.

Question 3. If the intensity of a sound wave is increased by a factor of 100, how much does the sound level in decibels (dB) increase?

  • a) \(10\ \mathrm{dB}\)
  • b) \(20\ \mathrm{dB}\)
  • c) \(50\ \mathrm{dB}\)
  • d) \(100\ \mathrm{dB}\)

Correct Answer: b) \(20\ \mathrm{dB}\)

Reason: The sound level is \(\beta=10\log_{10}(I/I_0)\), so the change is

\[ \Delta\beta=10\log_{10}\!\left(\frac{I_2}{I_1}\right)=10\log_{10}(100)=10\times2=20\ \mathrm{dB} \]

Question 4. A source emits a sound of frequency \(500\ \mathrm{Hz}\). If the source moves towards a stationary observer with a speed of \(50\ \mathrm{m\,s^{-1}}\) (velocity of sound in air \(=340\ \mathrm{m\,s^{-1}}\)), what is the frequency heard by the observer?

  • a) \(525\ \mathrm{Hz}\)
  • b) \(540\ \mathrm{Hz}\)
  • c) \(580\ \mathrm{Hz}\)
  • d) \(620\ \mathrm{Hz}\)

Correct Answer: c) \(580\ \mathrm{Hz}\) (closest option)

Reason: For a source moving towards a stationary observer:

\[ f’=f\,\frac{v}{v-v_s}=500\times\frac{340}{340-50}=500\times\frac{340}{290}\approx586\ \mathrm{Hz} \]

The closest option to \(586\ \mathrm{Hz}\) is \(580\ \mathrm{Hz}\).

Question 5. An observer moves with a speed of \(20\ \mathrm{m\,s^{-1}}\) towards a stationary sound source of frequency \(400\ \mathrm{Hz}\). What frequency does the observer perceive? (Speed of sound in air \(=340\ \mathrm{m\,s^{-1}}\))

  • a) \(420\ \mathrm{Hz}\)
  • b) \(440\ \mathrm{Hz}\)
  • c) \(460\ \mathrm{Hz}\)
  • d) \(480\ \mathrm{Hz}\)

Correct Answer: a) \(420\ \mathrm{Hz}\) (closest option)

Reason: For an observer moving towards a stationary source:

\[ f’=f\,\frac{v+v_o}{v}=400\times\frac{340+20}{340}=400\times\frac{18}{17}\approx423.5\ \mathrm{Hz} \]

The closest option is \(420\ \mathrm{Hz}\).

Question 6. A train moving at \(30\ \mathrm{m\,s^{-1}}\) blows a whistle at \(600\ \mathrm{Hz}\). What frequency is heard by a stationary observer standing in front of the train? (Speed of sound in air \(=340\ \mathrm{m\,s^{-1}}\))

  • a) \(640\ \mathrm{Hz}\)
  • b) \(660\ \mathrm{Hz}\)
  • c) \(680\ \mathrm{Hz}\)
  • d) \(700\ \mathrm{Hz}\)

Correct Answer: b) \(660\ \mathrm{Hz}\)

Reason: The source (train) moves towards the stationary observer:

\[ f’=f\,\frac{v}{v-v_s}=600\times\frac{340}{310}\approx658\ \mathrm{Hz}\approx660\ \mathrm{Hz} \]

Question 7. Two trains approach each other, each moving at \(20\ \mathrm{m\,s^{-1}}\). One train blows a whistle of \(500\ \mathrm{Hz}\). What is the frequency heard by the driver of the other train? (Speed of sound in air \(=340\ \mathrm{m\,s^{-1}}\))

  • a) \(520\ \mathrm{Hz}\)
  • b) \(550\ \mathrm{Hz}\)
  • c) \(580\ \mathrm{Hz}\)
  • d) \(600\ \mathrm{Hz}\)

Correct Answer: b) \(550\ \mathrm{Hz}\) (closest option)

Reason: The source and the observer both move towards each other:

\[ f’=f\,\frac{v+v_o}{v-v_s}=500\times\frac{340+20}{340-20}=500\times\frac{360}{320}=562.5\ \mathrm{Hz} \]

The closest option is \(550\ \mathrm{Hz}\).

Question 8. A person hears a sound of frequency \(450\ \mathrm{Hz}\). If the source is moving away with a speed of \(30\ \mathrm{m\,s^{-1}}\), what was the original frequency of the sound? (Speed of sound in air \(=340\ \mathrm{m\,s^{-1}}\))

  • a) \(480\ \mathrm{Hz}\)
  • b) \(470\ \mathrm{Hz}\)
  • c) \(460\ \mathrm{Hz}\)
  • d) \(450\ \mathrm{Hz}\)

Correct Answer: a) \(480\ \mathrm{Hz}\) (closest option)

Reason: For a source moving away, \(f’=f\dfrac{v}{v+v_s}\):

\[ 450=f\times\frac{340}{370}\;\Rightarrow\;f=450\times\frac{370}{340}\approx490\ \mathrm{Hz} \]

The closest option is \(480\ \mathrm{Hz}\).

Question 9. A supersonic jet moves towards an observer at a speed of \(400\ \mathrm{m\,s^{-1}}\) while emitting a sound of frequency \(1000\ \mathrm{Hz}\). What frequency does the observer hear? (Speed of sound in air \(=340\ \mathrm{m\,s^{-1}}\))

  • a) \(2000\ \mathrm{Hz}\)
  • b) \(3000\ \mathrm{Hz}\)
  • c) \(4000\ \mathrm{Hz}\)
  • d) No sound is heard

Correct Answer: d) No sound is heard

Reason: The jet is supersonic (\(v_s=400\ \mathrm{m\,s^{-1}}>v=340\ \mathrm{m\,s^{-1}}\)), so it travels faster than the sound it emits. The sound waves pile up into a Mach cone (shock wave) behind the jet, and an observer in front of the approaching jet hears nothing until the jet has passed.

Question 10. A stationary observer hears a frequency of \(550\ \mathrm{Hz}\) from a moving car. If the actual frequency of the horn is \(500\ \mathrm{Hz}\), what is the speed of the car? (Speed of sound in air \(=340\ \mathrm{m\,s^{-1}}\))

  • a) \(30\ \mathrm{m\,s^{-1}}\)
  • b) \(40\ \mathrm{m\,s^{-1}}\)
  • c) \(50\ \mathrm{m\,s^{-1}}\)
  • d) \(60\ \mathrm{m\,s^{-1}}\)

Correct Answer: a) \(30\ \mathrm{m\,s^{-1}}\) (closest option)

Reason: The heard frequency is higher than the actual frequency, so the car is moving towards the observer:

\[ 550=500\times\frac{340}{340-v_s}\;\Rightarrow\;1.1(340-v_s)=340\;\Rightarrow\;v_s=\frac{34}{1.1}\approx30.9\ \mathrm{m\,s^{-1}} \]

The closest option is \(30\ \mathrm{m\,s^{-1}}\).

Question 11. A police radar emits a sound wave of frequency \(1000\ \mathrm{Hz}\) towards a moving car. If the car is approaching at \(30\ \mathrm{m\,s^{-1}}\) and the speed of sound is \(340\ \mathrm{m\,s^{-1}}\), what is the frequency of the sound received by the car?

  • a) \(1088\ \mathrm{Hz}\)
  • b) \(1094\ \mathrm{Hz}\)
  • c) \(1100\ \mathrm{Hz}\)
  • d) \(1110\ \mathrm{Hz}\)

Correct Answer: a) \(1088\ \mathrm{Hz}\)

Reason: The car acts as a moving observer approaching the source:

\[ f_1=f\,\frac{v+v_c}{v}=1000\times\frac{370}{340}\approx1088\ \mathrm{Hz} \]

Note: if the car then reflects this sound back, it acts as a moving source and the police receive \(f_2=1088\times\dfrac{340}{310}\approx1194\ \mathrm{Hz}\).

Question 12. A ship sends an ultrasonic wave of frequency \(40\ \mathrm{kHz}\) towards the seabed. If the reflected wave is received with a frequency shift of \(400\ \mathrm{Hz}\) due to the motion of the ship, what is the speed of the ship? (Speed of sound in seawater \(=1500\ \mathrm{m\,s^{-1}}\))

  • a) \(5\ \mathrm{m\,s^{-1}}\)
  • b) \(7.5\ \mathrm{m\,s^{-1}}\)
  • c) \(10\ \mathrm{m\,s^{-1}}\)
  • d) \(12.5\ \mathrm{m\,s^{-1}}\)

Correct Answer: b) \(7.5\ \mathrm{m\,s^{-1}}\)

Reason: The ship is first a moving source and then a moving observer of the wave reflected from the stationary seabed: \(f_2=f\dfrac{v+v_s}{v-v_s}\). For \(v_s\ll v\) the shift is

\[ \Delta f\approx f\,\frac{2v_s}{v}\;\Rightarrow\;400=40000\times\frac{2v_s}{1500}\;\Rightarrow\;v_s=7.5\ \mathrm{m\,s^{-1}} \]

3. Numerical and Subjective Questions

Question 1. Pressure amplitude and the intensity of a sound wave.

a) Define pressure amplitude. How does it relate to sound wave propagation?

Sound waves are longitudinal mechanical waves made of alternating compressions and rarefactions. The pressure amplitude \(\Delta P_{\max}\) is the maximum increase or decrease in the local pressure of the medium from its undisturbed equilibrium value.

As the wave propagates it transports energy. The intensity (energy per unit time per unit area) is proportional to the square of the pressure amplitude:

\[ I=\frac{(\Delta P_{\max})^2}{2\rho v} \]

b) A sound wave has a pressure amplitude of \(0.5\ \mathrm{Pa}\) and travels through air of density \(1.2\ \mathrm{kg\,m^{-3}}\) with a velocity of \(340\ \mathrm{m\,s^{-1}}\). Find the intensity of the wave.

Given: \(\Delta P_{\max}=0.5\ \mathrm{Pa},\ \rho=1.2\ \mathrm{kg\,m^{-3}},\ v=340\ \mathrm{m\,s^{-1}}\)
\[ I=\frac{(\Delta P_{\max})^2}{2\rho v}=\frac{(0.5)^2}{2\times1.2\times340}=\frac{0.25}{816}\approx3.06\times10^{-4}\ \mathrm{W\,m^{-2}} \]
Question 2. Intensity and loudness of sound, and the decibel scale.

a) Differentiate between intensity and loudness of sound. How are they measured?

IntensityLoudness
Objective, purely physical quantity: the acoustic energy passing through a unit area per unit time.Subjective, physiological sensation of how strong a sound seems to the listener.
Measured with acoustic instruments in watts per square metre (\(\mathrm{W\,m^{-2}}\)).Depends on the intensity and on the ear’s sensitivity to frequency; measured on a logarithmic scale in decibels (dB) or phons.

b) A sound of intensity \(1.0\times10^{-8}\ \mathrm{W\,m^{-2}}\) is increased to \(1.0\times10^{-5}\ \mathrm{W\,m^{-2}}\). By how many decibels has the loudness increased?

Given: \(I_1=1.0\times10^{-8}\ \mathrm{W\,m^{-2}},\ I_2=1.0\times10^{-5}\ \mathrm{W\,m^{-2}}\)
\[ \Delta\beta=10\log_{10}\!\left(\frac{I_2}{I_1}\right)=10\log_{10}\!\left(\frac{10^{-5}}{10^{-8}}\right)=10\log_{10}(10^3)=30\ \mathrm{dB} \]
Question 3. Pitch and quality of sound, and the intensity from a point source.

a) Explain the concepts of pitch and quality of sound. Give examples of how different instruments produce sound with different quality.

Pitch: pitch is the brain’s interpretation of the fundamental frequency of a sound wave. A high-frequency sound (a bird chirping) has a high pitch and a low-frequency sound (a bass drum) has a low pitch.

Quality (timbre): quality is the characteristic that lets a listener distinguish two sounds of the same pitch and loudness. It depends on the waveform and the harmonic content (overtones) of the sound.

Example: a flute and a violin playing the same note (middle C, about 261 Hz) at the same loudness sound different. The flute gives a waveform close to a pure sine wave with few overtones, while the violin gives a complex waveform rich in overtones. The ear recognizes this mixture of overtones as the quality of the instrument.

b) A source of sound emits uniformly in all directions with a total power of \(2\ \mathrm{W}\). What is the intensity at a distance of \(3\ \mathrm{m}\)?

Given: \(P=2\ \mathrm{W},\ r=3\ \mathrm{m}\)

For a spherical wave:

\[ I=\frac{P}{4\pi r^2}=\frac{2}{4\pi(3)^2}=\frac{1}{18\pi}\approx0.0177\ \mathrm{W\,m^{-2}} \]
Question 4. Variation of the intensity of sound with distance from the source.

a) How does the intensity of sound vary with distance from the source?

For a point source radiating in three dimensions without reflection or absorption, the intensity is inversely proportional to the square of the distance from the source (inverse square law):

\[ I\propto\frac{1}{r^2} \]

The total power spreads over the surface of an expanding sphere of area \(4\pi r^2\).

b) The sound from a speaker has an intensity of \(1.0\times10^{-6}\ \mathrm{W\,m^{-2}}\) at a distance of \(2\ \mathrm{m}\). What is the intensity at \(4\ \mathrm{m}\) from the source?

Given: \(I_1=1.0\times10^{-6}\ \mathrm{W\,m^{-2}},\ r_1=2\ \mathrm{m},\ r_2=4\ \mathrm{m}\)
\[ I_2=I_1\left(\frac{r_1}{r_2}\right)^2=(1.0\times10^{-6})\left(\frac24\right)^2=2.5\times10^{-7}\ \mathrm{W\,m^{-2}} \]
Question 5. Loudness, the Doppler effect and an observer moving towards a source.

a) Discuss how the human ear perceives loudness and its logarithmic relation with intensity.

The ear can respond to an enormous range of intensities, trillions of times between the faintest and the loudest sounds. According to the Weber-Fechner law, the sensation of loudness is proportional not to the intensity but to the logarithm of the intensity. To make a sound seem about twice as loud, the intensity must be increased roughly ten times. For this reason the sound level is measured on a logarithmic scale:

\[ \beta=10\log_{10}\!\left(\frac{I}{I_0}\right)\ \mathrm{dB} \]

b) State and explain the Doppler effect. What factors influence the observed frequency?

Doppler effect: the apparent change in the frequency (or wavelength) of a wave received by an observer because of relative motion between the source and the observer. For example, the pitch of a siren is higher as an ambulance approaches and drops suddenly as it passes.

The observed frequency depends on three velocities: the velocity of the source \(v_s\), the velocity of the observer \(v_o\), and the velocity of sound \(v\) in the medium.

c) An observer moves towards a stationary source of frequency \(600\ \mathrm{Hz}\) with a speed of \(20\ \mathrm{m\,s^{-1}}\). What is the frequency heard? (\(v=340\ \mathrm{m\,s^{-1}}\))

Given: \(f=600\ \mathrm{Hz},\ v_o=20\ \mathrm{m\,s^{-1}},\ v=340\ \mathrm{m\,s^{-1}}\)
\[ f’=f\,\frac{v+v_o}{v}=600\times\frac{360}{340}=600\times\frac{18}{17}\approx635.3\ \mathrm{Hz} \]
Question 6. Derivation of the Doppler effect for a moving source and a moving observer.
Behind: wavelength longer(lower frequency)Ahead: wavelength shorter (higher frequency)source

Figure: Wavefronts of a source moving to the right

a) Derive the expression for the Doppler effect when (i) the source is moving and the observer is at rest, (ii) the observer is moving and the source is at rest, and (iii) both are moving.

Let \(v\) be the speed of sound, \(f\) the true frequency and \(\lambda=v/f\) the true wavelength.

(i) Source moving with speed \(v_s\), observer at rest

If the source moves towards the observer it follows its own waves. In one period \(T=1/f\) the source moves a distance \(v_sT\), so the wavelength in front of it is shortened:

\[ \lambda’=\lambda-v_sT=\frac{v}{f}-\frac{v_s}{f}=\frac{v-v_s}{f} \]

The observer hears

\[ f’=\frac{v}{\lambda’}=f\left(\frac{v}{v-v_s}\right) \]

(Use \(+v_s\) in the denominator if the source moves away.)

(ii) Observer moving with speed \(v_o\), source at rest

The wavelength \(\lambda=v/f\) is unchanged, but the observer moving towards the source meets the wavefronts at a relative speed \(v+v_o\):

\[ f’=\frac{v+v_o}{\lambda}=f\left(\frac{v+v_o}{v}\right) \]

(Use \(-v_o\) if the observer moves away.)

(iii) Both source and observer moving

The source motion changes the wavelength to \(\lambda’=\dfrac{v\mp v_s}{f}\) and the observer motion changes the relative speed to \(v\pm v_o\):

\[ f’=\frac{v\pm v_o}{v\mp v_s}\,f \]

The upper signs are used when they approach each other and the lower signs when they move apart.

b) Both the source and the observer move towards each other. The source emits sound of frequency \(440\ \mathrm{Hz}\) and moves at \(30\ \mathrm{m\,s^{-1}}\); the observer moves at \(20\ \mathrm{m\,s^{-1}}\). Find the apparent frequency. (\(v=340\ \mathrm{m\,s^{-1}}\))

Given: \(f=440\ \mathrm{Hz},\ v_s=30\ \mathrm{m\,s^{-1}},\ v_o=20\ \mathrm{m\,s^{-1}},\ v=340\ \mathrm{m\,s^{-1}}\)
\[ f’=f\,\frac{v+v_o}{v-v_s}=440\times\frac{340+20}{340-30}=440\times\frac{360}{310}=\frac{15840}{31}\approx511.0\ \mathrm{Hz} \]
Question 7. Applications of the Doppler effect, and a source moving away from an observer.

a) What is the significance of the Doppler effect in real life? Mention at least three applications.

The Doppler effect lets scientists and engineers measure the speed of distant or inaccessible objects just by analysing the waves they emit or reflect. Three applications:

(i) Medical imaging: Doppler ultrasound and echocardiograms measure the speed and direction of blood flow in the heart and blood vessels.
(ii) Radar: police radar guns and weather radar measure the speed of vehicles and storm clouds from the shift in the reflected waves.
(iii) Astronomy: the red shift or blue shift in the light of stars and galaxies shows how fast they are receding or approaching, and so how the universe is expanding.

b) A source of sound approaches a stationary observer. What happens to the frequency heard? Explain why.

The frequency heard increases (the pitch rises).

Each successive crest is emitted from a point closer to the observer than the previous one, so the distance between crests (the wavelength) decreases. The wave speed stays constant, so more crests reach the ear per second, and the observer hears a higher frequency.

c) A source emitting sound of frequency \(400\ \mathrm{Hz}\) moves away from an observer at rest with a speed of \(40\ \mathrm{m\,s^{-1}}\). What frequency is heard? (\(v=340\ \mathrm{m\,s^{-1}}\))

Given: \(f=400\ \mathrm{Hz},\ v_s=40\ \mathrm{m\,s^{-1}},\ v_o=0,\ v=340\ \mathrm{m\,s^{-1}}\)
\[ f’=f\,\frac{v}{v+v_s}=400\times\frac{340}{380}=\frac{13600}{38}\approx357.9\ \mathrm{Hz} \]
Question 8. Condition for no Doppler shift, and a whistle moving towards a stationary observer.

a) What is the condition under which no Doppler shift is observed even if there is relative motion?

No Doppler shift is observed when the relative motion does not change the distance between the source and the observer, that is, when there is no component of relative velocity along the line joining them. This happens, for example, when the source moves in a circle around a stationary observer: the wavefronts are neither compressed nor stretched as received by the observer.

b) A whistle of frequency \(500\ \mathrm{Hz}\) moves towards a stationary observer with a speed of \(30\ \mathrm{m\,s^{-1}}\). The speed of sound is \(340\ \mathrm{m\,s^{-1}}\). What frequency does the observer hear?

Given: \(f=500\ \mathrm{Hz},\ v_s=30\ \mathrm{m\,s^{-1}},\ v_o=0,\ v=340\ \mathrm{m\,s^{-1}}\)
\[ f’=f\,\frac{v}{v-v_s}=500\times\frac{340}{310}=\frac{17000}{31}\approx548.4\ \mathrm{Hz} \]

4. View PDF Notes

i. Handwritten Notes

ii. Typed PDF Notes

5. Frequently Asked Questions

What are acoustic phenomena in Class 12 Physics?
Acoustic phenomena is the NEB Class 12 Physics chapter on sound waves. It covers the pressure amplitude of sound waves, the characteristics of sound (intensity, loudness, quality and pitch) and the Doppler effect.
What topics are included in Chapter 9 Acoustic Phenomena of NEB Class 12 Physics?
The chapter (5 hours, part of the unit Waves and Optics) covers: 9.1 sound waves and pressure amplitude; 9.2 characteristics of sound: intensity, loudness, quality and pitch; 9.3 the Doppler effect.
What is the pressure amplitude of a sound wave?
Pressure amplitude is the maximum increase or decrease in the local pressure of the medium from its equilibrium value as a sound wave passes. It is related to the intensity by I = (ΔP_max)²/(2ρv), where ρ is the density of the medium and v is the speed of sound, so the intensity is proportional to the square of the pressure amplitude.
What is the intensity of sound and what is its SI unit?
Intensity is the energy carried by the wave per unit time across a unit area perpendicular to its direction of travel, I = P/A. Its SI unit is the watt per square metre (W m⁻²). For a sound wave, I = ½ρvω²A², so the intensity is proportional to the square of the amplitude.
How does the intensity of sound vary with distance from a source?
For a point source radiating uniformly in all directions, I = P/(4πr²), so the intensity is inversely proportional to the square of the distance (inverse square law). Doubling the distance reduces the intensity to one quarter.
What is the difference between intensity and loudness?
Intensity is an objective physical quantity measured in W m⁻². Loudness is the subjective sensation of how strong a sound seems, and it depends on the intensity and on the sensitivity of the ear to the frequency. Loudness is measured on a logarithmic scale in decibels (dB) or phons.
What is the decibel scale for sound level?
The sound level is β = 10 log₁₀(I/I₀) decibels, where I₀ is the threshold of hearing, about 10⁻¹² W m⁻². A change in intensity from I₁ to I₂ changes the level by Δβ = 10 log₁₀(I₂/I₁); for example, increasing the intensity 100 times raises the level by 20 dB.
What is the pitch of a sound?
Pitch is the sensation by which a sound is judged to be high or low. It depends mainly on the frequency of the sound: a high-frequency sound has a high pitch (a bird chirping) and a low-frequency sound has a low pitch (a bass drum).
What is the quality (timbre) of sound?
Quality or timbre is the characteristic that lets us tell apart two sounds of the same pitch and loudness. It depends on the harmonic content, the number and relative intensities of the overtones present. A flute and a violin playing the same note sound different because their overtones differ.
What is the Doppler effect?
The Doppler effect is the apparent change in the frequency of a wave heard by an observer because of relative motion between the source and the observer. For example, the pitch of an ambulance siren rises as it approaches and falls as it moves away.
What is the formula for the Doppler effect of sound?
The apparent frequency is f′ = f (v ± v_o)/(v ∓ v_s), where v is the speed of sound, v_o the speed of the observer and v_s the speed of the source. Use +v_o in the numerator when the observer moves towards the source and −v_s in the denominator when the source moves towards the observer; use the opposite signs when moving apart.
How do you find the frequency heard when a source moves towards a stationary observer?
Use f′ = f v/(v − v_s). For example, a 500 Hz whistle moving at 30 m s⁻¹ towards a stationary observer with v = 340 m s⁻¹ is heard at 500 × 340/310 ≈ 548.4 Hz. If the source moves away, use f′ = f v/(v + v_s).
What are the applications of the Doppler effect?
Applications include Doppler ultrasound and echocardiography to measure blood flow, police radar guns and weather radar to measure the speed of vehicles and storms, sonar for ships, and astronomy, where the red shift and blue shift of light show how fast stars and galaxies move.
When is no Doppler shift observed?
No Doppler shift is observed when there is no relative motion along the line joining the source and the observer, for example when the source moves in a circle around a stationary observer. A supersonic source moving faster than sound produces a shock wave (Mach cone), and an observer ahead of it hears nothing until it has passed.
Where can I get Class 12 Physics Chapter 9 Acoustic Phenomena notes in PDF?
This page includes handwritten notes and typed PDF notes for Acoustic Phenomena that you can view online, along with solved multiple choice questions and numericals.

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