IOE entrance Physics · Chapter 4
Waves and Sound
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36 questions in 4 syllabus topics · 15 are 2-mark questions.
4.1 Wave motion
9 questions
1. A wave is described by y = 0.05 sin(20πt − 4πx), where x and y are in metres and t in seconds. The speed of the wave is
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Wave speed = ω/k.
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Answer: A. 5 m/s
Compare with y = A sin(ωt − kx): ω = 20π rad/s and k = 4π rad/m. Wave speed v = ω/k = 20π/4π = 5 m/s.
2. In a stationary wave, the distance between a node and the nearest antinode is
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Nodes are λ/2 apart; where is the antinode?
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Answer: A. λ/4
Adjacent nodes are λ/2 apart and an antinode lies midway between them, so node-to-antinode distance is λ/4.
3. A sound wave of frequency 100 Hz travels with speed 400 m/s. The phase difference between two points on the wave whose path difference is 0.5 m is
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Δφ = (2π/λ)Δx.
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Answer: A. π/4
λ = v/f = 400/100 = 4 m. Phase difference = (2π/λ) × path difference = (2π/4) × 0.5 = π/4.
4. Which of the following statements about a stationary wave is correct?
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Think about what happens to phase as you cross a node.
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Answer: D. All particles between two adjacent nodes vibrate in the same phase
In a stationary wave the amplitude varies from zero (node) to maximum (antinode), no net energy flows, and particles within one loop move in phase while adjacent loops are in opposite phase.
5. A stationary wave is given by y = 4 sin(πx/15) cos(96πt), where x and y are in cm and t in s. The amplitude of vibration of a particle at x = 5 cm is
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The amplitude depends on x through the sine factor.
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Answer: D. 2√3 cm
Amplitude at position x is 4 sin(πx/15). At x = 5 cm: 4 sin(π/3) = 4 × (√3/2) = 2√3 cm.
6. Transverse mechanical waves cannot propagate through
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Which medium cannot resist a shearing deformation?
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Answer: B. gases
Transverse waves need a medium with shear rigidity (or surface tension at a surface). A gas has no rigidity, so only longitudinal waves travel through it.
7. For the wave y = 0.02 sin(100t − 5x) in SI units, the ratio of the maximum particle speed to the wave speed is
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Particle speed max = Aω; wave speed = ω/k.
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Answer: B. 0.1
Maximum particle speed = Aω = 0.02 × 100 = 2 m/s. Wave speed = ω/k = 100/5 = 20 m/s. Ratio = 2/20 = 0.1 (equal to Ak).
8. Which equation represents a progressive wave travelling in the negative x-direction?
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Keep the phase constant and see how x changes with t.
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Answer: D. y = A sin(ωt + kx)
A wave of the form f(ωt + kx) keeps constant phase when x decreases as t increases, so it travels towards −x. sin(ωt − kx) and sin(kx − ωt) move towards +x; A sin ωt cos kx is a stationary wave.
9. Two waves y₁ = a sin(ωt − kx) and y₂ = a sin(ωt + kx) superpose. The antinodes of the resulting stationary wave are located at (n = 0, 1, 2, …)
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Add the sines and find where the amplitude factor is maximum.
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Answer: C. x = nλ/2
y₁ + y₂ = 2a cos kx sin ωt. Antinodes occur where |cos kx| = 1, i.e. kx = nπ, so x = nλ/2. Nodes are at (2n + 1)λ/4.
4.2 Speed of sound
9 questions
10. Laplace corrected Newton's formula for the speed of sound in a gas by assuming that the compressions and rarefactions are
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Is there time for heat to flow during a rapid compression?
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Answer: C. adiabatic
Compressions and rarefactions are too fast for heat exchange, so the process is adiabatic and the bulk modulus is γP instead of P, giving v = √(γP/ρ).
11. The speed of sound in air at 27°C is v. At what temperature will the speed of sound in air be 2v?
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Speed of sound is proportional to √T in kelvin.
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Answer: B. 927°C
v ∝ √T. For speed to double, T must become 4 times: 4 × 300 K = 1200 K = 927°C.
12. If the pressure of air is doubled while its temperature is kept constant, the speed of sound in it
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How does density change when P changes at constant T?
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Answer: A. remains unchanged
v = √(γP/ρ). At constant temperature P/ρ is constant (Boyle's law), so the speed does not depend on pressure.
13. At the same temperature and pressure, sound travels in humid air
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Compare the molar mass of water vapour with that of air.
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Answer: B. faster than in dry air, because humid air is less dense
Water vapour (M = 18) is lighter than air (M ≈ 29), so humid air has lower density and v = √(γP/ρ) increases.
14. The ratio of the speed of sound in hydrogen to that in oxygen at the same temperature is (both diatomic)
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v ∝ 1/√M for gases with the same γ and T.
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Answer: A. 4
v = √(γRT/M) with the same γ, so v_H₂/v_O₂ = √(M_O₂/M_H₂) = √(32/2) = 4.
15. Newton's formula gives the speed of sound in air as 280 m/s. Taking γ = 1.44 for air, the speed according to Laplace's formula is
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The two formulae differ by a factor √γ.
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Answer: C. 336 m/s
Laplace speed = √γ × Newton speed = √1.44 × 280 = 1.2 × 280 = 336 m/s.
16. A person puts his ear on one end of a steel rail 1020 m long, and a hammer strikes the other end. If the speed of sound is 340 m/s in air and 5100 m/s in steel, the time interval between the two sounds heard is
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Find the travel time in each medium separately.
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Answer: D. 2.8 s
Time through air = 1020/340 = 3.0 s; time through steel = 1020/5100 = 0.2 s. Interval = 3.0 − 0.2 = 2.8 s.
17. For a diatomic gas (γ = 1.4), the ratio of the speed of sound to the rms speed of its molecules at the same temperature is
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Write both speeds in terms of RT/M.
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Answer: D. √(7/15)
v_sound = √(γRT/M) and v_rms = √(3RT/M), so ratio = √(γ/3) = √(1.4/3) = √(7/15).
18. Young's modulus of steel is 2 × 10¹¹ Pa and its density is 8000 kg/m³. The speed of longitudinal sound waves in a steel rod is
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In a thin solid rod, v = √(Y/ρ).
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Answer: B. 5000 m/s
v = √(Y/ρ) = √(2 × 10¹¹/8000) = √(2.5 × 10⁷) = 5 × 10³ m/s.
4.3 Waves in pipes and strings
9 questions
19. A pipe closed at one end and open at the other produces
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Fit a node and an antinode at the two ends.
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Answer: D. only odd harmonics
With a node at the closed end and an antinode at the open end, L = (2n − 1)λ/4, so frequencies are f, 3f, 5f, … — only odd harmonics.
20. A pipe closed at one end is 85 cm long. If the speed of sound is 340 m/s, the frequency of its first overtone is (neglect end correction)
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The first overtone of a closed pipe is not the second harmonic.
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Answer: D. 300 Hz
Fundamental f = v/4L = 340/(4 × 0.85) = 100 Hz. A closed pipe has only odd harmonics, so the first overtone is the third harmonic = 300 Hz.
21. The ratio of the fundamental frequency of an open pipe to that of a closed pipe of the same length is
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Compare v/2L and v/4L.
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Answer: A. 2 : 1
Open pipe: f = v/2L; closed pipe: f = v/4L. Ratio = (v/2L)/(v/4L) = 2 : 1.
22. In a resonance tube experiment with a tuning fork of 500 Hz, the first and second resonances occur at air column lengths of 16 cm and 50 cm. The end correction is
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The difference of the two lengths is λ/2.
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Answer: C. 1 cm
l₁ + e = λ/4 and l₂ + e = 3λ/4. So λ/2 = l₂ − l₁ = 34 cm, λ/4 = 17 cm, and e = 17 − 16 = 1 cm. (Also v = 0.68 × 500 = 340 m/s.)
23. The end correction at one open end of a pipe of internal radius r is approximately
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The antinode forms a little beyond the open end; the correction is a fraction of the radius, found empirically.
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Answer: A. 0.6r
The antinode forms slightly outside the open end, at about 0.6r (= 0.3 × diameter) beyond it.
24. The tension in a stretched string is increased by 44% keeping its length and mass unchanged. Its fundamental frequency increases by
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Frequency varies as the square root of tension.
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Answer: C. 20%
f ∝ √T, so f'/f = √1.44 = 1.2, an increase of 20%.
25. A wire 1 m long has a mass of 10 g and is stretched with a tension of 400 N. The frequency of its fundamental mode is
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Find the wave speed on the wire first, then use f = v/2L.
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Answer: B. 100 Hz
μ = 0.01 kg/1 m = 0.01 kg/m. v = √(T/μ) = √(400/0.01) = 200 m/s. f = v/2L = 200/2 = 100 Hz.
26. At the closed end of a closed organ pipe there is always a
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Can air move at a rigid wall? Displacement and pressure extremes are swapped.
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Answer: D. displacement node and pressure antinode
Air molecules cannot move at the rigid closed end, so displacement is zero (node) and the pressure variation is maximum (antinode).
27. The fundamental frequency of a closed pipe equals the frequency of the first overtone of an open pipe of length 60 cm. The length of the closed pipe is (neglect end corrections)
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The first overtone of an open pipe is its second harmonic.
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Answer: C. 15 cm
First overtone of open pipe = 2 × v/(2 × 0.6) = v/0.6. Closed pipe fundamental = v/(4l). Equating: 4l = 0.6 m, l = 0.15 m = 15 cm.
4.4 Acoustic phenomena and Doppler effect
9 questions
28. If the intensity of a sound is increased 100 times, its intensity level increases by
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Use L = 10 log₁₀(I/I₀).
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Answer: B. 20 dB
ΔL = 10 log₁₀(I₂/I₁) = 10 log₁₀100 = 10 × 2 = 20 dB.
29. Two sounds of the same frequency in the same medium have intensity levels of 60 dB and 40 dB. The ratio of their pressure amplitudes is
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Intensity is proportional to the square of pressure amplitude.
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Answer: A. 10
A 20 dB difference means an intensity ratio of 10² = 100. Since I ∝ (pressure amplitude)², the pressure-amplitude ratio is √100 = 10.
30. The quality (timbre) of a musical note depends on
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What differs between a violin and a flute playing the same note?
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Answer: C. the number and relative intensities of the overtones present
Two instruments playing the same pitch at the same loudness sound different because their waveforms contain different harmonics with different relative intensities.
31. Which of the following frequencies lies in the ultrasonic range?
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Audible range is about 20 Hz to 20 kHz.
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Answer: D. 30 kHz
Ultrasonic waves have frequencies above about 20 kHz, the upper limit of human hearing. 15 Hz is infrasonic; 2 kHz and 18 kHz are audible.
32. A source of sound of frequency 900 Hz moves towards a stationary observer at 34 m/s. If the speed of sound is 340 m/s, the frequency heard by the observer is
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For a source approaching, the denominator is (v − vₛ).
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Answer: D. 1000 Hz
f' = f × v/(v − vₛ) = 900 × 340/306 = 900 × 10/9 = 1000 Hz.
33. Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. The number of beats heard per second is
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Beat frequency is the difference of the two frequencies.
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Answer: D. 4
Beat frequency = |f₁ − f₂| = 260 − 256 = 4 per second.
34. A tuning fork A produces 5 beats per second with a fork of 300 Hz. When a little wax is stuck on A, it produces 3 beats per second with the same fork. The original frequency of A is
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Loading a fork decreases its frequency.
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Answer: A. 305 Hz
A is 295 or 305 Hz. Loading with wax lowers A's frequency. If A = 305 Hz, lowering it brings it closer to 300 Hz, so beats decrease (5 → 3), as observed. If A were 295 Hz, beats would increase. So A = 305 Hz.
35. The pitch of a sound is determined mainly by its
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Which property distinguishes a shrill sound from a grave one?
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Answer: C. frequency
Pitch is the sensation of how shrill a sound is, and it rises with frequency. Amplitude decides loudness and waveform decides quality.
36. A car sounding a horn of frequency 500 Hz moves at 20 m/s towards a cyclist who is moving at 20 m/s towards the car. If the speed of sound is 340 m/s, the frequency heard by the cyclist is
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Observer approaching adds to v in the numerator; source approaching subtracts from v in the denominator.
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Answer: D. 562.5 Hz
Both approach: f' = f(v + v₀)/(v − vₛ) = 500 × 360/320 = 562.5 Hz.