Sound Is Longitudinal
Sound in air travels through compressions and rarefactions. Particles vibrate parallel to propagation while pressure disturbance moves forward.
Class 11 Physics notes covering sound waves, longitudinal nature, pressure variation, speed of sound, Laplace correction, intensity, loudness, quality, pitch, numericals and PYQs.
Sound in air travels through compressions and rarefactions. Particles vibrate parallel to propagation while pressure disturbance moves forward.
Newton used isothermal compression and got a low value. Laplace used adiabatic compression and obtained v = √(γP/ρ).
Intensity is physical, loudness is subjective, pitch depends on frequency, and quality depends on waveform and harmonics.
Sound is one of the most important mechanical waves in Class 11 Physics because it connects wave motion with daily life, music, hearing, echo, ultrasound, communication and measurement of distance.
A sound wave is a longitudinal mechanical wave. It needs a material medium such as air, water, steel or wood. It cannot travel through vacuum because there are no particles to pass compressions and rarefactions.
Real-life example: when a bell rings, the metal surface vibrates and pushes nearby air layers. These air layers create alternate high-pressure and low-pressure regions that move outward as sound.
Exam perspective: NEET and CBSE focus on definitions, speed of sound and temperature dependence. JEE often asks Newton-Laplace correction, pressure-density phase, intensity and conceptual traps.
Sound waves are produced by vibrating bodies. The vibration creates periodic compression and expansion of the surrounding medium, and this disturbance travels outward carrying energy.
Production of sound requires a source, a medium and a receiver. The source vibrates, the medium transfers energy layer by layer, and the receiver such as an ear or microphone detects pressure variation.
Propagation happens due to elasticity and inertia. Elasticity restores compressed layers, while inertia carries particles beyond equilibrium, creating the next rarefaction and compression.
Mathematically, the speed of sound in a medium depends on elasticity divided by inertia. That is why the general form is v = √(elastic property / density).
Real-life example: sound travels faster in steel than in air because steel is much more elastic for mechanical disturbances, even though it is denser.
Sound in air is longitudinal because air particles vibrate parallel to the direction of propagation. If the sound travels to the right, air particles oscillate to and fro along the same line.
Compression is a region where particles are closer than normal, pressure is high and density is high. Rarefaction is a region where particles are farther apart, pressure is low and density is low.
The distance between two consecutive compressions or two consecutive rarefactions is one wavelength. One compression followed by one rarefaction forms half a wavelength.
Real-life example: push and pull a slinky spring along its length. Dense coil regions and loose coil regions move along the spring just like compressions and rarefactions in air.
Common mistake: do not call compressions crests and rarefactions troughs in a particle diagram. Use crest/trough mainly for displacement or pressure graphs.
A sound wave can be described as pressure variation, density variation or particle displacement variation. Pressure and density are maximum at compression and minimum at rarefaction.
Pressure variation is not the same as displacement variation. In a progressive sound wave, pressure variation and particle displacement are out of phase by π/2 in the standard description.
When displacement of particles is maximum, pressure variation is zero; when displacement changes most rapidly with position, pressure variation is maximum.
Real-life example: a microphone mainly responds to pressure variation of air, while a small dust particle in the air shows local displacement due to the sound wave.
Exam trap: pressure maximum occurs at compression, but particle displacement need not be maximum there.
Newton assumed that compressions and rarefactions in sound propagation are isothermal, meaning temperature remains constant. His formula was v = √(P/ρ), where P is pressure and ρ is density.
Using air pressure and density, Newton's formula gives about 280 m s−1, which is much lower than the experimental value near 331 m s−1 at 0°C.
The historical problem was important: the wave theory was correct, but the thermodynamic assumption was wrong. Sound compressions and rarefactions happen very rapidly, so there is not enough time for heat exchange.
The general idea remains valuable: speed of sound depends on elasticity and density. Greater elasticity increases speed; greater density decreases speed if elasticity is unchanged.
Real-life example: sound reaches faster through railway tracks than through air because the solid track transmits elastic disturbance more effectively.
Laplace corrected Newton's formula by saying that sound propagation in a gas is adiabatic, not isothermal. Correct formula: v = √(γP/ρ).
γ is the ratio of specific heats: γ = Cp/Cv. For air, γ is approximately 1.4.
Compression and rarefaction are adiabatic because they occur very rapidly. Heat does not get enough time to flow from compressed hot regions to rarefied cool regions.
Laplace correction works because adiabatic bulk modulus of a gas is γP, not P. Replacing P by γP increases the predicted speed from Newton's value to the experimental value.
JEE-level point: the correction is not a small empirical adjustment; it comes from thermodynamics of rapid pressure changes.
Temperature: in air, speed approximately follows v = v0 + 0.61T, where v0 is speed at 0°C and T is temperature in °C.
As temperature increases, gas molecules move faster and pressure response becomes quicker, so speed of sound increases. At 0°C speed is about 331 m s−1; at 20°C it is about 343 m s−1.
Humidity increases the speed of sound in air because water vapour makes moist air less dense than dry air at the same temperature and pressure.
Pressure alone has almost no effect on speed of sound in an ideal gas at constant temperature because P/ρ remains constant.
Nature of medium matters strongly: sound generally travels fastest in solids, slower in liquids and slowest in gases, due to different elasticity and density.
Intensity is sound power transmitted per unit area perpendicular to the direction of propagation. Formula: I = P/A, where P is power and A is area.
SI unit of intensity is W m−2. Intensity is an objective physical quantity and can be measured by instruments.
For a point source spreading sound uniformly, intensity decreases with distance because the same power spreads over a larger spherical area.
Real-life example: as you move away from a loudspeaker, sound intensity decreases. Doubling distance from a point source ideally reduces intensity to one-fourth.
Common mistake: intensity and loudness are not identical. Intensity is physical power per area; loudness is human sensation.
Loudness is the subjective sensation of how strong a sound appears to the human ear. It depends mainly on intensity, but also on frequency and ear sensitivity.
A sound of same intensity may not feel equally loud at all frequencies because the human ear is more sensitive around the speech-frequency range.
Real-life example: a high-frequency whistle and a low-frequency hum may have the same measured intensity but may not be perceived equally loud.
Examination perspective: if a question asks physical quantity, answer intensity; if it asks sensation perceived by ear, answer loudness.
Memory trick: intensity belongs to instruments; loudness belongs to ears.
Quality, also called timbre, is the property by which two sounds of same pitch and loudness can be distinguished.
It depends on waveform, harmonics and overtones present in the sound. Different instruments produce different mixtures of harmonics.
Real-life example: a flute and a violin can play the same note at the same loudness, but we can identify them because their waveforms and harmonics are different.
Musical instruments have a fundamental frequency plus harmonics. The relative strength of these harmonics decides the quality of sound.
Common trap: quality is not frequency alone. Frequency decides pitch; harmonic composition decides quality.
Pitch is the sensation that lets us classify sound as high or low. It depends mainly on frequency.
Higher frequency gives higher pitch; lower frequency gives lower pitch. A whistle has high pitch, while a drum has low pitch.
Human hearing typically covers about 20 Hz to 20,000 Hz. Frequencies below 20 Hz are infrasonic, and above 20 kHz are ultrasonic.
Real-life example: tightening a guitar string increases frequency and therefore pitch.
Common mistake: pitch and loudness are different. A sound can be high-pitched but soft, or low-pitched but loud.
v = √(γP/ρ)γ is Cp/Cv, P is pressure and ρ is density. Use for sound in gases.
v = √(E/ρ)E is elastic modulus and ρ is density. This expresses elasticity versus inertia.
I = P/AI is intensity, P is power and A is area normal to propagation.
v = v0 + 0.61TApproximate speed in air at T°C, with v0 near 331 m s−1.
v = fλConnects speed, frequency and wavelength of sound.
d = vt/2Echo time is round trip time, so one-way distance is half of vt.
| Point | Newton | Laplace |
|---|---|---|
| Assumption | Compression and rarefaction are isothermal. | Compression and rarefaction are adiabatic. |
| Formula | v = √(P/ρ) | v = √(γP/ρ) |
| Prediction in air | About 280 m s−1, too small. | About 331 m s−1 at 0°C, close to experiment. |
| Reason | Assumes heat exchange during compression. | Recognises rapid sound propagation leaves no time for heat exchange. |
| Exam importance | Historical error and conceptual trap. | Correct NEET/JEE formula for gases. |
These diagrams use black lines with red labels and arrows, matching board-work style for Class 11 Physics.
If Sound Waves or Wave Motion is not clear and you are looking for a Physics Tutor, contact Kumar Sir.
This numerical bank avoids repeated data-only duplicates and covers temperature, Laplace correction, Newton formula, intensity, echo, pitch, wavelength, medium comparison, humidity reasoning and sound perception.
Given: T = 20°C, v0 = 331
Formula: v = v0 + 0.61T
Solution: v = 331 + 0.61 × 20 = 343.2 m s−1
Final Answer: 343.2 m s−1
Given: T = 30°C
Formula: v = 331 + 0.61T
Solution: v = 331 + 18.3 = 349.3 m s−1
Final Answer: 349.3 m s−1
Given: f = 500 Hz, v = 340
Formula: λ = v/f
Solution: λ = 340/500 = 0.68 m
Final Answer: 0.68 m
Given: λ = 2 m, f = 170 Hz
Formula: v = fλ
Solution: v = 170 × 2 = 340 m s−1
Final Answer: 340 m s−1
Given: P = 20 W, A = 5 m2
Formula: I = P/A
Solution: I = 20/5 = 4 W m−2
Final Answer: 4 W m−2
Given: t = 2 s, v = 340
Formula: d = vt/2
Solution: d = 340 × 2/2 = 340 m
Final Answer: 340 m
Given: v = 343.2, v0 = 331
Formula: T = (v - v0)/0.61
Solution: T = (343.2 - 331)/0.61 = 20°C
Final Answer: 20°C
Given: λ = 0.5 m, v = 350
Formula: f = v/λ
Solution: f = 350/0.5 = 700 Hz, higher than 350 Hz so higher pitch
Final Answer: 700 Hz, higher pitch
Given: P = 100 W, A = 200 m2
Formula: I = P/A
Solution: I = 100/200 = 0.5 W m−2
Final Answer: 0.5 W m−2
Given: I' = 4I
Formula: I ∝ A2
Solution: A'/A = √4 = 2
Final Answer: Amplitude doubles
Given: P = 1.01 × 105, ρ = 1.29
Formula: v = √(P/ρ)
Solution: v = √(1.01 × 105/1.29) = 279.8 m s−1
Final Answer: 279.8 m s−1
Given: γ = 1.4, P = 1.01 × 105, ρ = 1.29
Formula: v = √(γP/ρ)
Solution: v = √(1.4 × 1.01 × 105/1.29) = 331.0 m s−1
Final Answer: 331 m s−1
Given: vN = 280, γ = 1.4
Formula: vL = √γ vN
Solution: vL = √1.4 × 280 = 331.3 m s−1
Final Answer: 331.3 m s−1
Given: T = 40°C
Formula: v = 331 + 0.61T
Solution: v = 331 + 24.4 = 355.4 m s−1
Final Answer: 355.4 m s−1
Given: ΔT = 20°C
Formula: Δv = 0.61ΔT
Solution: Δv = 0.61 × 20 = 12.2 m s−1
Final Answer: 12.2 m s−1
Given: I = 8, A = 3
Formula: P = IA
Solution: P = 8 × 3 = 24 W
Final Answer: 24 W
Given: r doubles
Formula: I ∝ 1/r2
Solution: I' = I/22 = I/4
Final Answer: I/4
Given: f = 1000 Hz, λ = 0.34 m
Formula: v = fλ
Solution: v = 340 m s−1; 1000 Hz is audible
Final Answer: 340 m s−1, audible
Given: t = 0.04 s, v = 340
Formula: d = vt/2
Solution: d = 340 × 0.04/2 = 6.8 m
Final Answer: 6.8 m
Given: t = 0.6 s, v = 1500
Formula: d = vt/2
Solution: d = 1500 × 0.6/2 = 450 m
Final Answer: 450 m
Given: f doubles, v constant
Formula: λ = v/f
Solution: Wavelength halves; pitch increases
Final Answer: Wavelength halves, pitch higher
Given: Same f, same intensity, different waveform
Formula: Quality depends on waveform
Solution: Quality or timbre differs
Final Answer: Quality/timbre
Given: γ = 1.4
Formula: vL/vN = √γ
Solution: ratio = √1.4 = 1.183
Final Answer: 1.183
Given: T constant, ideal gas
Formula: P/ρ constant
Solution: Speed remains unchanged
Final Answer: No change
Given: γ = 5/3, P = 105, ρ = 1.2
Formula: v = √(γP/ρ)
Solution: v = √((5/3)×105/1.2) = 372.7 m s−1
Final Answer: 372.7 m s−1
Given: v = 1400, ρ = 1000
Formula: E = ρv2
Solution: E = 1000 × 14002 = 1.96 × 109 Pa
Final Answer: 1.96 × 109 Pa
Given: E same, ρ1:ρ2 = 1:4
Formula: v ∝ 1/√ρ
Solution: v1:v2 = 2:1
Final Answer: 2:1
Given: ρ same, E1:E2 = 9:16
Formula: v ∝ √E
Solution: v1:v2 = 3:4
Final Answer: 3:4
Given: P = 80 W, r = 2 m
Formula: I = P/(4πr2)
Solution: I = 80/(16π) = 5/π W m−2
Final Answer: 5/π W m−2
Given: r' = 3r
Formula: I ∝ 1/r2
Solution: I' = I/9
Final Answer: I/9
Given: v = 360
Formula: T = (v - 331)/0.61
Solution: T = 29/0.61 = 47.54°C
Final Answer: 47.5°C
Given: λ = 5 mm = 0.005 m, v = 340
Formula: f = v/λ
Solution: f = 340/0.005 = 68000 Hz
Final Answer: 68 kHz
Given: I/I0 = 100
Formula: I ∝ A2
Solution: A/A0 = √100 = 10
Final Answer: 10
Given: E = 0.6 J, t = 10 s, A = 0.02
Formula: P = E/t, I = P/A
Solution: P = 0.06 W; I = 0.06/0.02 = 3 W m−2
Final Answer: 3 W m−2
Given: Newton: P, Laplace: γP
Formula: vL/vN = √γ
Solution: Since γ > 1, Laplace speed is larger
Final Answer: Laplace speed is √γ times Newton speed
Given: vN = 300, vL = 354
Formula: γ = (vL/vN)2
Solution: γ = (354/300)2 = 1.3924
Final Answer: 1.39
Given: γ1 = 1.4, γ2 = 1.67
Formula: v ∝ √γ
Solution: v2/v1 = √(1.67/1.4) = 1.092
Final Answer: 1.092
Given: v = 5000, ρ = 8000
Formula: E = ρv2
Solution: E = 8000 × 25 × 106 = 2.0 × 1011 Pa
Final Answer: 2.0 × 1011 Pa
Given: Pressure antinode in sound
Formula: Pressure and density variations are in phase
Solution: Density variation is maximum there
Final Answer: Maximum density variation
Given: v1 = 330, v2 = 345, f = 690
Formula: λ = v/f
Solution: λ1 = 0.478 m; λ2 = 0.500 m; increase = 0.022 m
Final Answer: 0.022 m increase
Given: r1 = 5, r2 = 20
Formula: I ∝ 1/r2
Solution: I2/I1 = (5/20)2 = 1/16
Final Answer: 1/16
Given: I1:I2 = 16:1
Formula: I ∝ A2
Solution: A1:A2 = 4:1
Final Answer: 4:1
Given: T = 27°C, distance = 1000 m
Formula: v = 331 + 0.61T, t = s/v
Solution: v = 347.47; t = 1000/347.47 = 2.88 s
Final Answer: 2.88 s
Given: T = 30°C, echo time = 3 s
Formula: v = 331 + 0.61T, d = vt/2
Solution: v = 349.3; d = 349.3 × 3/2 = 523.95 m
Final Answer: 524 m
Given: E = 12 J, t = 4 s, A = 6
Formula: P = E/t, I = P/A
Solution: P = 3 W; I = 3/6 = 0.5 W m−2
Final Answer: 0.5 W m−2
Given: Frequency doubles, amplitude same
Formula: Pitch depends on frequency
Solution: Pitch increases; loudness may remain nearly same
Final Answer: Pitch increases
Given: Loudness and pitch are different
Formula: Loudness depends intensity; pitch depends frequency
Solution: A loud drum can have low pitch and a soft whistle can have high pitch
Final Answer: Statement is false
Given: Progressive sound wave
Formula: Pressure and displacement differ by π/2
Solution: Pressure variation is maximum where displacement gradient is maximum
Final Answer: Phase difference π/2
Given: Moist air has lower density than dry air at same T and P
Formula: v = √(γP/ρ)
Solution: Lower density gives higher speed
Final Answer: Humidity increases speed
Given: P changes with ρ proportionally
Formula: v = √(γP/ρ)
Solution: P/ρ remains constant at fixed T
Final Answer: Speed unchanged
Given: Same pitch, same loudness
Formula: Quality depends on waveform and harmonics
Solution: Their quality or timbre differs
Final Answer: Quality/timbre
Given: f = 40 kHz
Formula: Human range about 20 Hz to 20 kHz
Solution: 40 kHz is above hearing range
Final Answer: Not audible
Given: Amplitude increases, frequency same
Formula: Loudness depends on amplitude/intensity
Solution: Loudness increases; pitch remains same
Final Answer: Loudness increases
Given: Waveform changes
Formula: Quality depends on waveform
Solution: Quality/timbre changes
Final Answer: Quality changes
Includes CBSE, NEET, JEE Main, JEE Advanced, IB, ICSE, IGCSE, A-Level, assertion-reason, true-false, case-study, reasoning and conceptual questions.
Because it requires a material medium for propagation.
Longitudinal mechanical wave.
Compression is high pressure-density region; rarefaction is low pressure-density region.
No, because sound needs a material medium.
v = √(P/ρ).
v = √(γP/ρ).
Power transmitted per unit area normal to propagation.
W m−2.
It assumed isothermal changes, but sound propagation in gas is adiabatic.
Ratio of specific heats, Cp/Cv.
It increases with temperature.
v = v0 + 0.61T.
No, because P/ρ remains constant.
It increases sound speed in air.
Intensity is objective.
Frequency.
They happen too fast for heat exchange.
Laplace speed is √γ times Newton speed.
v = √(E/ρ) or v = √(elastic modulus/density).
π/2.
Pressure and density variations are in phase.
v = √(B/ρ).
It uses adiabatic bulk modulus γP instead of isothermal modulus P.
Speed increases as √γ.
Solids have much larger elastic modulus.
Yes, if harmonic composition differs.
Loudness is subjective sensation; intensity is measurable power per area.
Pitch is perception; frequency is physical quantity mainly deciding pitch.
Quality of sound determined by waveform and harmonics.
A frequency component above the fundamental.
Compressions.
Rarefactions.
About 20 Hz to 20 kHz.
Sound with frequency above 20 kHz.
Bell in evacuated jar becomes faint or inaudible.
Intensity is larger near the source.
Frequency and pitch increase.
Different harmonics and waveform.
Intensity is proportional to amplitude squared.
I ∝ 1/r2.
Adiabatic process.
Adiabatic bulk modulus γP.
Both true and reason explains assertion.
Both true and reason explains assertion.
Both true and reason explains assertion.
Assertion false, reason true.
False.
False.
True.
True.
True.
Sound requires a material medium.
d = vt/2.
Different timbre due to different harmonics.
v = v0 + 0.61T.
It depends on human ear sensitivity, not only physical intensity.
It is measurable power per unit area.
Yes, pitch and loudness are different.
Yes, if their waveforms/harmonics differ.
They represent different variables and are phase shifted.
Medium particles oscillate locally while disturbance propagates.
If Sound Waves or Wave Motion is not clear and you are looking for a Physics Tutor, contact Kumar Sir.