NDA General Studies · Physics · 47 questions · 2017 to 2026

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 (4)4.2 The simple pendulum (9)4.3 Waves and the nature of sound (11)4.4 The speed of sound (7)4.5 Characteristics of sound: pitch, loudness, range and beats (9)4.6 Reflection of sound: echo, reverberation, ultrasound and SONAR (7)

4.1 Periodic motion and simple harmonic motion

1. NDA 2021 (I)
Which one of the following statements is true for a simple harmonic oscillator?
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.

2. NDA 2018 (I)
If T is the time period of an oscillating pendulum, which one of the following statements is NOT correct?
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).

3. NDA 2017 (I)
The following figure shows displacement versus time curve for a particle executing simple harmonic motion:
[figure]
Which one of the following statements is correct?
0 2 4 6 8 10 Time (s) Displacement
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.

4. NDA 2017 (I)
Which one of the following is the correct relation between frequency f and angular frequency ω?
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

5. NDA 2026 (I)
A pendulum of length L oscillates with an angular amplitude of θ = 60° and time period T. Let T0 = 2π⁠√(Lg)⁠ be the time period for small angle of oscillations, where g is the acceleration due to gravity. If air resistance is negligibly small and the string remains straight, then which one of the following is correct?
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.

6. NDA 2025 (I)
The length of a simple pendulum is increased four times to its previous value while the mass is doubled. What is the ratio of the new and previous time period of the pendulum?
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).

7. NDA 2024 (I)
Which one of the following statements regarding simple pendulum is correct?
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.

8. NDA 2022 (II)
A simple pendulum having bob of mass m and length of string l has time period of T. If the mass of the bob is doubled and the length of the string is halved, then the time period of this pendulum will be
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).

9. NDA 2022 (I)
The time period of a 1 m long pendulum approximates to
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).

10. NDA 2019 (II)
A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g/⁠2. The time period of the second pendulum would be:
(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).

11. NDA 2018 (II)
The time period of oscillation of a simple pendulum having length L and mass of the bob m is given as T. If the length of the pendulum is increased to 4L and the mass of the bob is increased to 2m, then which one of the following is the new time period of oscillation?
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.

12. NDA 2018 (I)
Which one of the following is an example of the force of gravity of the earth acting on a vibrating pendulum bob?
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).

13. NDA 2017 (I)
The time period of a simple pendulum made using a thin copper wire of length L is T. Suppose the temperature of the room in which this simple pendulum is placed increases by 30° C, what will be the effect on the time period of the pendulum?
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

14. NDA 2025 (I)
The frequency (f), wavelength (λ) and speed (v) of a sound wave are related as
Answer and solution

Answer: (d) v = λf

Method. wave relation

v = λf (NS10-A60, Eq. 10.2; KP14-A37, Eq. 14.12).

15. NDA 2024 (I)
Which one of the following does not apply to sound waves in fluids?
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).

16. NDA 2023 (II)
Which one of the following statements is true for sound waves propagating in air?
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).

17. NDA 2022 (I)
A sound wave has a frequency of 1 kHz and wavelength 50 cm. How long will it take to travel 1 km?
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.

18. NDA 2021 (II)
Which among the following is true for propagation of sound waves?
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).

19. NDA 2021 (I)
Which one of the following cannot be the unit of frequency of a sound wave?
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).

20. NDA 2020 (I)
Which of the following statements is NOT correct regarding the travel of sound waves?
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).

21. NDA 2020 (I)
Which one of the following statements about sound is NOT correct?
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).

22. NDA 2018 (II)
Statement I:
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).

23. NDA 2018 (I)
Which of the following statements about electromagnetic waves, sound waves and water waves is/⁠are correct?
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).

24. NDA 2017 (II)
Which one of the following waves does not belong to the category of the other three?
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

25. NDA 2026 (II)
Consider the following statements in respect of sound waves:
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.

26. NDA 2026 (I)
A very large container consists of an ideal gas. The speed of sound in the gas is x. When the pressure of the gas is doubled while keeping the temperature constant, the speed of sound now becomes y. What is the ratio of x to y?
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.

27. NDA 2023 (II)
The flash of lightning is seen before the thunderstorm is heard. It verifies that
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).

28. NDA 2023 (I)
Which one among the following is true for the speed of sound in a given medium?
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.

29. NDA 2022 (II)
Which one of the following statements about the speed of sound waves is not correct?
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).

30. NDA 2019 (I)
At 20°C, the speed of sound in water is approximately
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).

31. NDA 2017 (I)
The speed of a body that has Mach number more than 1 is
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

32. NDA 2023 (I)
Which one of the following statements is not true for a flute, a musical instrument?
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.

33. NDA 2022 (II)
When the pitch of sound increases, which one of the following increases?
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.

34. NDA 2022 (II)
The amplitude of sound waves is measured in the units of
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.

35. NDA 2021 (I)
‘Beats’ is a phenomenon that occurs when frequencies of two harmonic waves are
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.

36. NDA 2019 (II)
The loudness of sound depends upon the
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.

37. NDA 2019 (II)
Compared to audible sound waves, ultrasound waves have
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.

38. NDA 2018 (II)
Statement I:
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).

39. NDA 2018 (II)
The frequency of ultrasound waves is
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).

40. NDA 2018 (I)
Which one of the following frequency ranges is sensitive to human ears?
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

41. NDA 2025 (II)
SONAR stands for
Answer and solution

Answer: (c) Sound Navigation and Ranging

Method. SONAR

SONAR = Sound Navigation And Ranging (NS10-A96).

42. NDA 2025 (I)
We hear an echo due to
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).

43. NDA 2022 (II)
SONAR is a device that is used to measure the distance of underwater objects by a ship. Which of the following types of waves does it use for this purpose?
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).

44. NDA 2021 (II)
Reverberation is a phenomenon associated with a
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).

45. NDA 2021 (I)
The sound created in a big hall persists because of the repeated reflections. The phenomenon is called
Answer and solution

Answer: (a) Reverberation.

Method. reverberation

Persistence of sound in a hall due to repeated reflections is reverberation (NS10-A90, NS10-A91).

46. NDA 2017 (II)
Bats detect obstacles in their path by receiving the reflected
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.

47. NDA 2017 (II)
Which one of the following statements is not correct?
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).

Take a free NDA paperBrowse exam prep

← Properties of matter and heatLight →