NDA Oscillations, waves and sound previous year questions with solutions
Oscillations, waves and sound is a small, fading topic of the General Studies paper, about one question in sixty-five. Its one-fact sound questions have thinned to one or two a sitting, now on the speed of sound and the pendulum. Every question UPSC has set on it since 2017, solved. Grouped by the method each needs, newest first; open a question for its answer and worked solution.
4.1 Periodic motion and simple harmonic motion
Answer and solution
Answer: (b) Force acting is directly proportional to the displacement from the mean position and is in opposite direction.
Method. force law of SHM
In SHM the force is F = -kx: proportional to displacement and opposite to it (KP13-A49, Eqs. 13.13-13.14). Acceleration a = -ω²x is not constant, and velocity is periodic.
Answer and solution
Answer: (a) The motion repeats after time T only once
Method. periodicity of motion
Periodic motion repeats after T, and after every nT (n = 1, 2, 3, …); T is the least such time (KP13-A76 'least time, nT′). So 'repeats after T only once' is wrong: it repeats again and again. (d) true: period is independent of amplitude only for small angles (KP13-A81).
[figure]
Which one of the following statements is correct?
Answer and solution
Answer: (c) Phase of the oscillating particle is same at t = 3 s and t = 7 s.
Method. phase repeats after one period
From the graph the curve starts at the origin and repeats every 4 s (crests near 1, 5, 9 s; troughs near 3, 7 s). Phase is equal at instants differing by a whole number of periods, 4 s, 8 s, … (KP13-A31, KP13-A14).
(a) 3 − 1 = 2 s, half a period: opposite phase.
(b) 8 − 2 = 6 s = 1.5 T: no.
(c) 7 − 3 = 4 s = T: same phase (both troughs). Correct.
(d) 10 − 4 = 6 s = 1.5 T: no.
Answer and solution
Answer: (b) ω = 2πf
Method. angular frequency and frequency
ω = 2π/T and f = 1/T, so ω = 2πf (KP13-A33, Eqs. 13.5-13.7).
4.2 The simple pendulum
Answer and solution
Answer: (a) T will be slightly greater than T0.
Method. large-amplitude pendulum
At θ0 = 60° sin θ < θ, so the restoring torque is weaker than the small-angle approximation says, and the period is longer: T ≈ T0(1 + θ0²/16 + …) = T0(1 + 0.068) ≈ 1.07 T0 (exact elliptic integral gives 1.073 T0). T is independent of mass (KP13-A66, KP13-A81 for the small-angle limit). So T > T0 slightly.
Answer and solution
Answer: (c) 2 : 1
Method. T ∝ √L
T = 2π√(L/g) is independent of mass. L -> 4L gives T′/T = √4 = 2, ratio 2 : 1 (KP13-A69).
Simple pendulum has a time period independent of amplitude:
Answer and solution
Answer: (d) only for small amplitudes because then the net force on its bob is proportional to its displacement.
Method. pendulum and small angles
Restoring torque -mgL sin θ; for small θ, sin θ ≈ θ, so the restoring force is proportional to displacement and the motion is SHM with T = 2π√(L/g) independent of amplitude (KP13-A66, KP13-A68, KP13-A81). For large amplitude sin θ ≠ θ, so not independent. The force is not independent of displacement.
Answer and solution
Answer: (b) T/√2
Method. T = 2π√(l/g)
Mass does not affect T. Length halved: T′ = 2π√((l/2)/g) = T/√2 (KP13-A69).
Answer and solution
Answer: (c) 2 s
Method. T = 2π√(L/g)
T = 2π√(1/10) = 1.99 s (g = 10 m/s²; 2.01 s with g = 9.8). About 2 s: the 'seconds pendulum' (KP13-A70, Example 13.8).
(in terms of period T of the first pendulum)
Answer and solution
Answer: (a) √2 T
Method. T proportional to 1/√g
T = 2π√(L/g); mass does not matter and L is the same. T2/T1 = √(g/(g/2)) = √2, so T2 = √2 T (KP13-A69).
Answer and solution
Answer: (b) 2T
Method. T = 2π√(L/g), independent of mass
T = 2π√(L/g) (KP13-A69). Mass does not enter. L -> 4L gives T′ = 2π√(4L/g) = 2T. Mass 2m makes no difference.
Answer and solution
Answer: (c) Restoring force
Method. restoring force on a pendulum
The tangential component of gravity, mg sin θ, always acts towards the mean position and is the restoring force of the pendulum (KP13-A64, KP13-A65).
Answer and solution
Answer: (a) T will increase slightly
Method. thermal expansion lengthens the pendulum
T = 2π√(L/g) (KP13-A69). Copper expands on heating, so L increases by a tiny fraction (ΔL/L = αΔT ~ 1.7e-5 x 30 ~ 5e-4), hence T increases by half of that, ~0.03%: slightly. Not 2 times.
4.3 Waves and the nature of sound
Answer and solution
Answer: (d) v = λf
Method. wave relation
v = λf (NS10-A60, Eq. 10.2; KP14-A37, Eq. 14.12).
Answer and solution
Answer: (c) They are transverse
Method. sound in fluids
In fluids (liquids and gases) sound is longitudinal, not transverse, because fluids cannot sustain shear (KP14-A17). The others hold: sound carries energy, needs a medium, and is faster in liquids than gases (KP14-A49).
Answer and solution
Answer: (b) Sound is a mechanical wave and longitudinal in nature
Method. nature of sound
Sound in air is a mechanical wave (needs a medium) and longitudinal (NS10-A30, NS10-A31; KP14-A09).
Answer and solution
Answer: (d) 2 s
Method. v = fλ, t = d/v
v = fλ = 1000 Hz x 0.50 m = 500 m/s (NS10-A60). t = 1000 m / 500 m/s = 2 s.
Answer and solution
Answer: (b) Sound cannot travel in vacuum and it is a longitudinal wave in air.
Method. nature of sound
Sound needs a medium, so it cannot travel in vacuum (NS10-A17), and in air it is a longitudinal wave of compressions and rarefactions (NS10-A30, KP14-A09).
Answer and solution
Answer: (a) dB
Method. units of frequency
Frequency is in Hz = s-1; min-1 (per minute) is also a valid unit (cycles per minute). dB is a unit of sound level, not of frequency (NS10-A75, NS10-A47).
Answer and solution
Answer: (d) Sound waves can travel through vacuum
Method. sound needs a medium
Sound travels through solids, liquids and gases (steel, water, air) but not through vacuum (NS10-A14, NS10-A17).
Answer and solution
Answer: (b) Sound waves are transverse waves
Method. nature of sound waves
(a) true: ~340 m/s against 3e8 m/s. (b) false: sound in air and fluids is longitudinal (NS10-A30, KP14-A09). (c) true. (d) true: humid air is less dense, so speed rises with humidity (NS10-A62). Not correct: (b).
Sound wave cannot propagate in vacuum.
Statement II:
Sound waves are elastic waves and require a medium to propagate.
Answer and solution
Answer: (a) Both the statements are individually true and Statement II is the correct explanation of Statement I
Method. sound needs a medium
Statement I true (bell jar experiment, NS10-A16, NS10-A17). Statement II true: sound waves are elastic (mechanical) waves that propagate by particle-to-particle compression and rarefaction, which needs a medium (NS10-A31, KP14-A03). II explains I: (a).
1. They exhibit reflection
2. They carry energy
3. They exert pressure
4. They can travel in vacuum
Select the correct answer using the code given below:
Answer and solution
Answer: (a) 1, 2 and 3
Method. properties of waves
1: all three reflect (NS10-A84; KP14-A59). 2: all carry energy (KP14-A02; NS10-A36). 3: all exert pressure (radiation pressure of EM waves; sound pressure; water waves exert force on surfaces). 4: only EM waves travel in vacuum; sound and water waves need a medium (KP14-A03, KP14-A04). So 1, 2 and 3 only is true of all three kinds: (a).
Answer and solution
Answer: (d) Sound waves
Method. mechanical vs electromagnetic waves
X-rays, microwaves and radio waves are electromagnetic (travel in vacuum at c). Sound is a mechanical, longitudinal wave needing a medium (NS10-A31, NS10-A34; KP14-A03, KP14-A04).
4.4 The speed of sound
I. The speed of sound waves decreases with an increase in temperature if the atmospheric pressure remains the same.
II. The speed of sound waves increases with increasing humidity if the atmospheric pressure remains the same.
III. The speed of sound waves is not affected by a change in atmospheric pressure, provided the temperature is kept constant.
Which of the statements given above are correct?
Answer and solution
Answer: (b) II and III only
Method. factors affecting the speed of sound
I false: speed rises with temperature (NS10-A62). II true: moist air is less dense, so speed rises with humidity (NS10-A62). III true: at constant T, v = √(γP/ρ) is independent of P(ρ ∝ P). So II and III only.
Answer and solution
Answer: (a) 1
Method. speed of sound in an ideal gas
v = √(γP/ρ) = √(γRT/M) for an ideal gas (KP14-A52, Laplace). At constant T, doubling P doubles ρ, so P/ρ is unchanged and v is unchanged: x/y = 1.
Answer and solution
Answer: (b) light travels much faster than sound
Method. speed of light vs sound
Light (3e8 m/s) is far faster than sound (~340 m/s), so the flash reaches us first (NS10-A65, Example 10.3, lightning distance).
Answer and solution
Answer: (a) Speed of sound remains same at all frequencies
Method. speed of sound independent of frequency
In a given medium the speed of sound depends on the medium (and temperature), not on frequency; changing f changes λ, v = fλ stays constant (NS10-A63). So it is the same at all frequencies (non-dispersive). (d) is false since v is constant, higher λ means lower f only.
Answer and solution
Answer: (b) The speed of sound waves in air decreases with increase in temperature.
Method. speed of sound
(a) true: steel ~5,000 m/s against water ~1,500 m/s. (c) true: speed in air rises with temperature (v ∝ √T; ~331 m/s at 0°C to ~344 m/s at 22°C; NS10-A62). So (b) is false. (d) true (NS10-A61).
Answer and solution
Answer: (c) 1500 m/s
Method. speed of sound in water
Speed of sound in water is about 1,480-1,500 m/s at room temperature (NCERT Class 11 Table 14.1, KP14-A48; NS10-A67, NS10-A61: liquids faster than gases (~340 m/s) but slower than solids).
Answer and solution
Answer: (a) supersonic
Method. Mach number = speed / speed of sound
Mach number M = v/v_sound. M > 1 means the body moves faster than sound: supersonic; M < 1 is subsonic. (NCERT Class 9 Science ch 10, NS10-A29: sonic boom from supersonic bodies; standard definition of Mach number.) Options c, d are particular speeds, not a regime.
4.5 Characteristics of sound: pitch, loudness, range and beats
Answer and solution
Answer: (d) Sound comes from a vibrating column of air inside as well as outside the flute.
Method. Sound production in an open-pipe wind instrument
Outside NCERT; by the physics of open pipes: the sound-producing resonance is the standing wave in the air column inside the flute tube (KP14-A73). The jet's momentum (flow strength) sets loudness and its travel time to the edge sets the pitch, consistent with (a) and (b). (c) is the standard statement. (d) claims the vibrating column extends outside, which is not true: outside air only carries the radiated sound. So (d) is the one not true.
Outside NCERT; (c) and (d) cannot both be true and (d) is chosen on that basis, with moderate confidence.
Answer and solution
Answer: (d) Frequency
Method. pitch and frequency
Pitch is the perceived highness of a note and is determined by frequency; higher frequency means higher pitch (NS10-A69). Wavelength decreases at higher frequency in the same medium.
Answer and solution
Answer: (a) pressure
Method. Amplitude of a sound wave
NCERT states that for sound the amplitude has the unit of density or pressure: the maximum variation of pressure (or density) from its mean (NS10-A54). So the nearest option is pressure.
Answer and solution
Answer: (d) nearly same.
Method. beats
Beats arise when two waves of nearly equal frequencies superpose; the amplitude pulsates at |f1 − f2| (KP14-A77, KP14-A79). Equal frequencies give no beats; far-apart frequencies give a beat rate too fast to be heard as beats.
Answer and solution
Answer: (b) amplitude of the sound waves.
Method. loudness and amplitude
Loudness depends on the amplitude (and the sensitivity of the ear); larger amplitude means more energy and more loudness (NS10-A73, NS10-A55). Frequency fixes pitch; speed is a property of the medium.
Answer and solution
Answer: (b) higher frequency.
Method. ultrasound vs audible sound
Ultrasound is defined by frequency above 20 kHz (NS10-A72). In the same medium both travel at the same speed (v independent of frequency, NS10-A63). Since λ = v/f, higher f means shorter wavelength. So only 'higher frequency' holds.
The pitch of the sound wave depends upon its frequency.
Statement II:
The loudness of the sound wave depends upon its amplitude.
Answer and solution
Answer: (b) Both the statements are individually true but Statement II is not the correct explanation of Statement I
Method. pitch and loudness
Statement I true: pitch is set by frequency (NS10-A69). Statement II true: loudness depends on amplitude (NS10-A73, NS10-A55). But loudness is not the explanation of pitch: they are independent properties. So both true, II not the explanation: (b).
Answer and solution
Answer: (d) greater than 20 kHz
Method. ultrasound range
Ultrasound is sound of frequency above 20 kHz, the upper limit of human hearing (NS10-A72).
Answer and solution
Answer: (b) 20–20,000 Hz
Method. audible range
The audible range of humans is about 20 Hz to 20,000 Hz (NS10-A71).
4.6 Reflection of sound: echo, reverberation, ultrasound and SONAR
Answer and solution
Answer: (c) Sound Navigation and Ranging
Method. SONAR
SONAR = Sound Navigation And Ranging (NS10-A96).
Answer and solution
Answer: (b) Reflection of sound waves
Method. echo
An echo is sound reflected from a distant surface and heard after at least 0.1 s (NS10-A85, NS10-A86).
Answer and solution
Answer: (c) Ultrasonic waves
Method. SONAR
SONAR (Sound Navigation And Ranging) sends ultrasonic pulses and times their echo; ultrasound is used because it is highly directional and short-wavelength, and audible sound would disturb and spread (NS10-A96).
Answer and solution
Answer: (b) multiple reflection of sound.
Method. reverberation
Reverberation is the persistence of sound caused by multiple reflections from walls, floor and ceiling (NS10-A90).
Answer and solution
Answer: (a) Reverberation.
Method. reverberation
Persistence of sound in a hall due to repeated reflections is reverberation (NS10-A90, NS10-A91).
Answer and solution
Answer: (b) Ultrasonic waves
Method. echolocation by bats
Bats emit ultrasonic pulses and detect the echoes reflected from obstacles and prey (NS10-A95, echolocation). Infrasonic is below 20 Hz; radio waves and microwaves are not used.
Answer and solution
Answer: (a) Ultrasonic waves cannot get reflected, refracted or absorbed.
Method. properties of ultrasound
Ultrasonic waves, like all sound waves, are reflected, refracted and absorbed (NS10-A84, NS10-A94; echo ranging and SONAR depend on reflection), so (a) is false. (b) flaw detection and (c) drilling hard materials by ultrasonic machining are standard uses (NS10-A94); (d) true, sound needs a medium (NS10-A17).