Original practice sets for Sound Waves are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
1. Sound waves in air are:
Explanation: Sound propagates through alternating compressions and rarefactions — particle motion is along the direction of wave travel (longitudinal).
2. Fundamental wave relation is:
Explanation: Wave speed = frequency × wavelength. v = fλ applies to all periodic waves.
3. Speed of sound in air at 0°C is approximately:
Explanation: Speed of sound in air at 0°C ≈ 332 m/s. At 25°C it's ≈ 343 m/s. For JEE, 330–332 m/s at 0°C is used.
4. Speed of sound in a gas is given by:
Explanation: Sound propagation is adiabatic. Newton's formula √(P/ρ) underestimates by factor √γ. Laplace corrected it to √(γP/ρ).
5. Intensity of a sound wave is proportional to:
Explanation: Intensity I ∝ A²f². For a given frequency, intensity is proportional to the square of amplitude.
6. Sound intensity level in decibels is defined as:
Explanation: β(dB) = 10 log₁₀(I/I₀) where I₀ = 10⁻¹² W/m² (threshold of hearing).
7. Speed of sound in air doubles when temperature (Kelvin) is:
Explanation: v ∝ √T. To double v, need 4T. E.g., from 300 K to 1200 K doubles speed of sound.
8. Speed of sound is greatest in:
Explanation: Sound speed: solids > liquids > gases. In steel ≈ 5000 m/s, in water ≈ 1500 m/s, in air ≈ 340 m/s. No sound in vacuum.
9. Threshold of pain (120 dB) is how many times louder than threshold of hearing (0 dB)?
Explanation: β = 10 log(I/I₀). At 120 dB: 120 = 10 log(I/I₀) → I/I₀ = 10¹². Intensity is 10¹² times higher.
10. Angular frequency ω of a sound wave with frequency f is:
Explanation: ω = 2πf (rad/s). In the wave equation y = A sin(ωt − kx), ω = 2πf and k = 2π/λ.
11. At what temperature does sound speed in air equal twice its value at 0°C?
Explanation: v ∝ √T. 2v₀ requires T = 4T₀ = 4×273 = 1092 K = 819°C.
12. Two sources of intensities I and 4I. Combined intensity level compared to I alone is higher by:
Explanation: Combined: I+4I = 5I. Δβ = 10 log(5I/I) = 10 log5 ≈ 10×0.699 = 7 dB. Actually: increase = 10 log(5) ≈ 7 dB. For 4I alone vs I: Δβ = 10 log4 ≈ 6 dB. The question compares I+4I vs I: 10log5 ≈ 7 dB.
13. At distance r from a point source, intensity I ∝ 1/r². If distance doubles, dB level decreases by:
Explanation: I₂ = I₁/4. Δβ = 10 log(I₂/I₁) = 10 log(1/4) = −6 dB. Level decreases by 6 dB when distance doubles.
14. Pressure wave equation in sound is p = p₀ sin(ωt − kx). Displacement wave lags pressure by:
Explanation: Displacement y = y₀ cos(ωt − kx) = y₀ sin(ωt − kx + π/2). Pressure wave is π/2 ahead of displacement — they are 90° out of phase.
15. Speed of sound in an ideal gas is √(γRT/M). For oxygen (M=32 g/mol, γ=1.4) at 300 K (R=8.314):
Explanation: v = √(1.4×8.314×300/0.032) = √(3491.88/0.032) = √(109121) ≈ 330 m/s.
16. Power of a speaker is 100 W. Intensity at 100 m assuming uniform spherical radiation:
Explanation: I = P/(4πr²) = 100/(4π×10⁴) = 100/125664 ≈ 7.96×10⁻⁴ ≈ 8×10⁻⁴ W/m².
17. If intensity increases by a factor of 1000, sound level increases by:
Explanation: Δβ = 10 log(1000) = 10×3 = 30 dB.
18. Speed of sound is higher in moist air than dry air because:
Explanation: Molar mass of water vapour (18 g/mol)
19. Bulk modulus of a gas relates to speed of sound by:
Explanation: Speed of sound in any medium = √(B/ρ) where B is elastic modulus (bulk modulus for gases). For adiabatic process, B = γP.
20. Loudness depends on:
Explanation: Loudness is a subjective perception. Equal-loudness contours show the ear is most sensitive around 3–4 kHz. Loudness depends on both the physical intensity and frequency of sound.
21. When two waves are superimposed, the resultant displacement is:
Explanation: Principle of superposition: y = y₁ + y₂. This leads to interference, beats, and standing waves.
22. Beat frequency when two tuning forks have frequencies f₁ and f₂ is:
Explanation: Beats arise from periodic constructive/destructive interference. Beat frequency = |f₁ − f₂|.
23. In a standing wave, nodes are points of:
Explanation: Nodes are formed where the two travelling waves cancel each other — always zero displacement amplitude.
24. At a rigid (closed) boundary, a sound pulse is reflected with:
Explanation: Reflection at a rigid boundary causes phase reversal (phase shift of π or 180°). At a free/open boundary, there is no phase change.
25. Tuning fork A (512 Hz) and B give 4 beats/second. B is loaded with wax and now gives 3 beats/second. Frequency of B before loading was:
Explanation: Loading with wax decreases B's frequency. Beats decrease from 4 to 3, so B was higher than A initially (516 Hz − 512 = 4). After loading: 516−Δ and beats = |512−(516−Δ)| decreases. B was 516 Hz.
26. Distance between consecutive nodes in a standing wave of wavelength λ is:
Explanation: Nodes occur at positions where path difference = nλ/2. Consecutive nodes are λ/2 apart.
27. Distance between a node and the nearest antinode is:
Explanation: Antinodes are midway between nodes. Node-to-antinode distance = λ/4.
28. Two waves of frequencies 198 Hz and 202 Hz are superimposed. Beat period is:
Explanation: Beat frequency = 202 − 198 = 4 Hz. Beat period = 1/4 = 0.25 s.
29. Minimum distance from a reflecting wall to hear an echo (assuming human ear can distinguish sounds 0.1 s apart, v_sound = 340 m/s):
Explanation: Echo heard after 0.1 s. Distance = v×t/2 = 340×0.1/2 = 17 m (sound travels to wall and back).
30. An open pipe resonates at 200 Hz. Second harmonic of this pipe is:
Explanation: Open pipe supports all harmonics: f_n = nf₁. f₂ = 2×200 = 400 Hz.
31. If a tuning fork of frequency 480 Hz gives 5 beats with a wire. When wire tension increases slightly, beat frequency decreases to 3. Wire frequency was:
Explanation: Increasing tension raises wire frequency. If beats decreased from 5 to 3, wire moved closer to fork — wire was below fork. Wire frequency = 480 − 5 = 475 Hz.
32. In a stationary wave, energy:
Explanation: Standing waves do not transfer energy. Energy oscillates between KE and PE locally at each point, with no net energy transport.
33. Two waves y₁ = A sin(ωt) and y₂ = A sin(ωt + 2π/3) superimpose. Resultant amplitude is:
Explanation: Resultant amplitude = 2A cos(Δφ/2) = 2A cos(π/3) = 2A × ½ = A.
34. Number of beats heard per second when two musical notes of 256 Hz and 260 Hz are played together on a piano:
Explanation: Beat frequency = |260 − 256| = 4 Hz = 4 beats per second.
35. A closed organ pipe and an open organ pipe are in unison at fundamental frequency. If open pipe is 30 cm long, closed pipe length is:
Explanation: Open pipe fundamental: f = v/(2L₀) = v/0.6. Closed pipe: f = v/(4L_c). Same f: v/0.6 = v/(4L_c) → L_c = 0.6/4 = 0.15 m = 15 cm.
36. Standing wave equation y = 2A sin(kx) cos(ωt) has antinodes at:
Explanation: Antinodes where sin(kx) = ±1 → kx = π/2, 3π/2, 5π/2,... = (2n+1)π/2 for n = 0,1,2,...
37. Reverberation time in acoustics is the time for sound intensity to fall by:
Explanation: Reverberation time (RT60): time for sound to decrease by 60 dB (factor of 10⁶ in intensity) after source stops. Defined by Sabine: T = 0.161V/A.
38. Beats are heard between two tuning forks only when:
Explanation: Beats require nearly equal frequencies. If difference is too large (>10–20 Hz for human ear), beat frequency is too fast to perceive as distinct pulses.
39. Two sound waves of amplitudes A and 2A with phase difference 60° superimpose. Resultant amplitude is:
Explanation: Resultant A_R = √(A₁² + A₂² + 2A₁A₂cosφ) = √(A² + 4A² + 2×A×2A×cos60°) = √(5A² + 2A²) = A√7.
40. In a closed pipe of length 1 m (v_sound = 340 m/s), frequencies of 3rd and 5th harmonics are:
Explanation: Closed pipe: f_n = nv/(4L) for odd n. f₁ = 340/4 = 85 Hz. f₃ = 3×85 = 255 Hz. f₅ = 5×85 = 425 Hz.
41. Fundamental frequency of an open pipe of length L is:
Explanation: Open pipe: both ends are antinodes. Fundamental has λ/2 = L → f₁ = v/(2L).
42. Fundamental frequency of a closed pipe of length L is:
Explanation: Closed pipe: one end node, one end antinode. Fundamental has λ/4 = L → f₁ = v/(4L).
43. Open organ pipe supports harmonics:
Explanation: Open pipe: f_n = nv/(2L) for n = 1, 2, 3,... All harmonics are present — richer tone.
44. Closed organ pipe supports harmonics:
Explanation: Closed pipe: f_n = nv/(4L) for n = 1, 3, 5,... Only odd harmonics. More hollow/nasal tone.
45. End correction in organ pipes accounts for:
Explanation: An open end acts as an antinode, but it forms slightly beyond the physical end. End correction e ≈ 0.6r (r = pipe radius) is added to each open end.
46. A resonance column experiment (closed pipe). Resonance lengths at fundamental and 3rd harmonic: L₁ and L₂. Speed of sound is:
Explanation: L₁ = λ/4, L₂ = 3λ/4. L₂ − L₁ = λ/2. v = fλ = 2f(L₂−L₁).
47. Open pipe of length L and closed pipe of length L/2 have fundamental frequencies in ratio:
Explanation: Open: v/(2L). Closed of length L/2: v/(4×L/2) = v/(2L). Same fundamental frequency — a useful comparison.
48. As temperature rises, pitch of an open pipe:
Explanation: f = v/(2L) ∝ v ∝ √T. As T increases, v increases, so f increases — pitch rises.
49. In a resonance column, first resonance occurs at 17.5 cm and second at 52.5 cm. The end correction implied by these idealised data is:
Explanation: $L_2-L_1=λ/2=35$ cm, so $λ=70$ cm. Since $L_1+e=λ/4=17.5$ cm, the implied end correction is 0 cm .
50. The 5th harmonic of an open pipe of length 0.5 m (v = 340 m/s) is:
Explanation: f_n = nv/(2L) = 5×340/(2×0.5) = 5×340 = 1700 Hz.
51. A closed pipe gives 5th overtone at 1700 Hz. Fundamental frequency is:
Explanation: Closed pipe: overtones are odd harmonics. 5th overtone = 11th harmonic (harmonics 1,3,5,7,9,11 — 5th overtone is n=11). f₁₁ = 11×f₁ = 1700 → f₁ ≈ 154.5 Hz.
52. Open pipe of physical length 30 cm (radius 1 cm, end correction e = 0.6r). Effective length:
Explanation: Two open ends: total correction = 2e = 2×0.6×1 = 1.2 cm. Effective length = 30 + 1.2 = 31.2 cm.
53. If an open pipe sounds 2nd harmonic simultaneously as a closed pipe sounds 3rd harmonic, ratio of their lengths is:
Explanation: Open 2nd harmonic: f = 2v/(2L_o) = v/L_o. Closed 3rd harmonic: f = 3v/(4L_c). Equal: v/L_o = 3v/(4L_c) → L_o/L_c = 4/3. Ratio L_o:L_c = 4:3.
54. Speed of sound measured by resonance column: L₁ = 25 cm (1st resonance) at 1000 Hz. Speed of sound is:
Explanation: At 1st resonance in closed pipe: L₁ = λ/4 (ignoring end correction). λ = 4×0.25 = 1 m. v = fλ = 1000×1... that gives 1000 m/s — too high. With end correction of ~0.5 cm: λ ≈ 4×0.25 = 1 m. But for L₁ = 25 cm to give v ≈ 333: need λ = 333/1000 = 0.333 m → L₁ = 8.33 cm. Adjusting for test: v = 4×f×L₁ = 4×1000×0.25 = 1000 m/s in plain calculation.
55. An open pipe is cut into two halves. Fundamental of each half compared to original:
Explanation: f₁ = v/(2L). When L → L/2, f₁ → v/(L) = 2f₁. Fundamental doubles — each half pipe sounds an octave higher.
56. Kundt's tube is used to measure:
Explanation: Kundt's tube: a rod clamped at its middle vibrates and drives standing waves in a tube with cork dust. The dust pattern reveals wavelength → speed.
57. A closed pipe of length 1 m (v = 300 m/s) is overblown to produce 3rd harmonic. This frequency is:
Explanation: Closed pipe 3rd harmonic: f₃ = 3v/(4L) = 3×300/4 = 225 Hz.
58. An open pipe sounds 200 Hz at 17°C. Assuming sound speed is proportional to √T, its frequency at 357°C is approximately:
Explanation: For a fixed pipe, $f∝v∝√T$. Thus $f_2=200\sqrt{630/290}≈ 295\text{ Hz} $.
59. If resonance column first resonates at 24 cm and second at 75 cm, the end correction is:
Explanation: L₂ − L₁ = λ/2 = 51 cm → λ = 102 cm. L₁ + e = λ/4 = 25.5 cm → e = 25.5 − 24 = 1.5 cm.
60. In an open pipe, if the pipe is stopped at one end (converted to closed), fundamental frequency becomes:
Explanation: Open pipe f₁ = v/(2L). Closed pipe of same L: f₁' = v/(4L) = f₁/2. Stopping reduces fundamental by half.
61. Doppler effect formula for sound: observer moving toward stationary source gives observed frequency:
Explanation: Observer moving toward source: f_observed = f₀(v+v₀)/v. More wavefronts hit observer per second.
62. Doppler effect formula: source moving toward stationary observer:
Explanation: Source approaching: wavefronts pile up in front → smaller effective wavelength → higher observed frequency. f = f₀v/(v−v_s).
63. When source and observer move away from each other, observed frequency:
Explanation: Recession stretches wavefronts → longer effective wavelength → lower frequency. Red-shift equivalent in sound.
64. A train moving at 72 km/h (v_sound = 340 m/s) sounds a horn at 500 Hz. Frequency heard by stationary observer in front is:
Explanation: v_s = 72 km/h = 20 m/s. Observer in front: f = 500 × 340/(340−20) = 500×340/320 = 531.25 ≈ 533 Hz.
65. Same train moving away from observer: frequency heard is:
Explanation: Source receding: f = 500×340/(340+20) = 500×340/360 = 472 Hz ≈ 471 Hz.
66. General Doppler formula for sound (both source and observer moving, taking toward positive):
Explanation: General formula: f = f₀(v ± v₀)/(v ∓ v_s). Use + for v₀ toward source, − for v_s toward observer. Sign convention varies; remember the physical meaning.
67. A bat emits 50,000 Hz and moves at 10 m/s toward a wall. Echo frequency heard by bat (v_sound = 330 m/s):
Explanation: Wall reflects as apparent source at f₁ = 50000×330/(330−10) = 51562 Hz. Then bat hears echo: f₂ = f₁×(330+10)/330 = 51562×340/330 ≈ 53125 Hz.
68. Doppler ultrasound in medicine detects:
Explanation: Ultrasound reflects off moving red blood cells. Doppler shift in reflected frequency reveals blood velocity and direction.
69. When source speed equals sound speed, Mach number is:
Explanation: Mach number M = v_source/v_sound. M = 1 means sonic speed — source moves with its own sound waves. A Mach cone is formed.
70. A supersonic aircraft (M = 2) produces a Mach cone. Half-angle of cone is:
Explanation: sin(θ) = v_sound/v_source = 1/M = 0.5 → θ = 30°. Half-angle of Mach cone = arcsin(1/M).
71. Observer moves at 10 m/s toward two stationary sources at the same location, emitting 1000 Hz and 990 Hz (v_sound = 340 m/s). The beat frequency heard is approximately:
Explanation: f₁ observed = 1000×(340+10)/340 = 1029 Hz. f₂ (stationary) = 990 Hz. Beats = 1029−990 = 39 Hz. Wait: observer vs second source also needs Doppler? If second source is also stationary: f₂ observed = 990×(340+10)/340 = 990×1.029 = 1019 Hz. Beats = 1029−1019 = 10 Hz. But if second source is at observer position... let me recalculate: both stationary sources, observer moving: f₁ = 1000×350/340 ≈ 1029 Hz, f₂ = 990×350/340 ≈ 1019 Hz. Beats = 10 Hz.
72. Doppler effect does NOT apply to:
Explanation: Doppler effect applies to all waves when there is relative motion between source and observer. A source at rest relative to observer shows no Doppler shift.
73. A car moves at 20 m/s toward a cliff. Driver sounds horn at 400 Hz (v = 340 m/s). Beats between original horn and echo heard by driver:
Explanation: Echo source moves toward driver (source = image of horn moving at 20 m/s toward cliff). Echo frequency at cliff: f₁ = 400×340/(340−20) = 425 Hz. Driver hears echo: f₂ = 425×(340+20)/340 = 425×360/340 = 450 Hz. Direct horn = 400 Hz. Beats = 450−400 = 50 Hz. Approximately 47 Hz.
74. The sonic boom from a supersonic aircraft is due to:
Explanation: Sonic boom is caused by the Mach cone (conical pressure wave) sweeping the ground as the aircraft passes. Observers hear the boom when the cone passes, not when aircraft is overhead.
75. A source emits f₀, moves in a circle at radius R and angular speed ω. Observer at the centre hears:
Explanation: When source moves in a circle with observer at centre, the radial velocity of source relative to observer is always zero. No Doppler shift — observer hears f₀ at all times.
76. Relativistic Doppler formula for light (source approaching): f_obs = f₀√((1+β)/(1−β)) where β = v/c. For v << c this reduces to:
Explanation: For v
77. The whistle of a stationary train is 1000 Hz. An observer approaches at 34 m/s (v_sound = 340 m/s). Observed frequency is:
Explanation: Observer moving toward stationary source: f = f₀(v+v₀)/v = 1000×(340+34)/340 = 1000×1.1 = 1100 Hz.
78. Crack of a rifle bullet (supersonic) is heard before the bang because:
Explanation: Supersonic bullet creates a Mach cone. The crack (N-wave shock) from the bullet passes the observer before the muzzle blast (which travels from behind). Hence crack is heard first.
79. A source moves toward observer at v/2 (half speed of sound). Observed frequency is:
Explanation: f = f₀v/(v−v_s) = f₀v/(v−v/2) = f₀v/(v/2) = 2f₀.
80. In Doppler radar, police detect vehicle speed because:
Explanation: Doppler radar: emitted frequency f₀ reflects off moving vehicle; reflected frequency is Doppler-shifted. Δf is proportional to vehicle speed, giving precise speed measurements.
81. In a Kundt's tube, cork dust forms heaps at nodes of the standing wave. If the frequency is 2 kHz and the distance between alternate heaps is 8 cm, speed of sound in the gas is:
Explanation: Distance between alternate heaps = λ/2 (consecutive nodes) × 2 = λ... Heaps form at nodes. Distance between successive node positions = λ/2. Alternate heaps = every other node = λ apart. 8 cm... If alternate heaps are λ/2 apart (consecutive nodes 8 cm): λ = 16 cm = 0.16 m. v = fλ = 2000×0.16 = 320 m/s.
82. Two trains approach each other. Train A (100 km/h) has horn at 600 Hz. Train B (50 km/h). v_sound = 340 m/s. Frequency of A's horn heard by B's passengers:
Explanation: v_A = 100/3.6 ≈ 27.8 m/s, v_B = 50/3.6 ≈ 13.9 m/s (toward source). f = 600 × (340+13.9)/(340−27.8) = 600 × 353.9/312.2 = 600 × 1.134 ≈ 680 Hz. Nearest: 676 Hz.
83. Sound diffracts around obstacles more than light because:
Explanation: Diffraction is significant when λ ≈ obstacle size. Sound: λ ∼ cm to m. Everyday objects (doors, buildings) are same scale. Light: λ ~ 500 nm — microscopic, so light doesn't bend around large everyday obstacles.
84. A tuning fork of frequency f is held near an open pipe of length L. As pipe length slowly increases from L₀, beats are first heard then become zero at length L₁ = v/(2f). This means:
Explanation: When L₁ = v/(2f), the pipe's fundamental matches the fork. Pipe vibrates in resonance with fork — single frequency, no beats. Beats disappear at resonance.
85. Infrasound (f < 20 Hz) is used to detect:
Explanation: Infrasound travels vast distances with little attenuation. Earthquakes, volcanoes, and severe storms generate infrasound. Arrays of microbarometers can detect eruptions thousands of km away.
86. Two waves y₁ = A sin(ω₁t) and y₂ = A sin(ω₂t) superimpose (ω₁ ≈ ω₂). Amplitude of resultant oscillates at:
Explanation: y = 2A cos((ω₁−ω₂)t/2) × sin((ω₁+ω₂)t/2). Amplitude envelope oscillates at (ω₁−ω₂)/2. Beat frequency (full amplitude cycle) = (ω₁−ω₂)/(2π) × 2π = ω₁−ω₂ in terms of angular frequency difference.
87. Acoustic intensity at 10 dB is:
Explanation: β = 10 log(I/I₀). 10 = 10 log(I/10⁻¹²) → log(I/10⁻¹²) = 1 → I = 10×10⁻¹² = 10⁻¹¹ W/m².
88. The half-angle of Mach cone satisfies sin θ = 1/M. For M = √2, θ equals:
Explanation: sin θ = 1/M = 1/√2 → θ = 45°.
89. In a musical instrument, a string coupled to a resonance box makes louder sound because:
Explanation: The box (sound board) is set into forced vibrations by the string. Its large surface area displaces much more air per unit time, radiating louder sound. It doesn't add energy — it transfers string energy to air more efficiently.
90. A source and observer both move at v_sound/2 in the same direction. Observed frequency:
Explanation: Same direction, same speed: relative velocity = 0. No Doppler shift. f = f₀×(v+v₀)/(v+v_s) = f₀×(v+v/2)/(v+v/2) = f₀.
91. In an open pipe, 4th overtone (5th harmonic) at 500 Hz. Pipe length (v = 340 m/s):
Explanation: f₅ = 5v/(2L) = 500 → L = 5×340/(2×500) = 1700/1000 = 1.7 m = 170 cm.
92. Most sound is reflected when sound passes from air to water because:
Explanation: Acoustic impedance Z = ρv. For air Z ≈ 415 Pa·s/m. For water Z ≈ 1.5×10⁶ Pa·s/m. Huge mismatch → almost total reflection. Transmission coefficient ∝ (Z₂−Z₁)²/(Z₂+Z₁)².
93. A sonometer wire of length 1 m, mass per unit length 1 g/m, under tension 100 N vibrates. Fundamental frequency:
Explanation: v = √(T/μ) = √(100/0.001) = √10⁵ = 316 m/s. f₁ = v/(2L) = 316/2 = 158 Hz.
94. The Hubble redshift (recession of distant galaxies) is analogous to Doppler effect for light. Recession of galaxy at speed 0.01c gives fractional frequency shift Δf/f of approximately:
Explanation: Δf/f ≈ v/c = 0.01c/c = 0.01 (classical Doppler for v
95. A Helmholtz resonator (cavity resonator) has resonant frequency f = (v/2π)√(A/(VL)) where A = neck area, V = cavity volume, L = neck length. To lower the resonant frequency:
Explanation: f ∝ √(A/VL). To decrease f: increase V (larger cavity) or decrease A (narrower neck) or increase L. Both V and A terms work as stated.
96. Group velocity v_g and phase velocity v_p for a non-dispersive medium (like sound in open air) are:
Explanation: In a non-dispersive medium, all frequency components travel at the same speed. Phase velocity = group velocity = wave speed. Dispersion occurs in waveguides or dense media where v depends on frequency.