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Oscillations: questions with answers

Class 11 Physics, Chapter 13 (NCERT). 15 questions by concept, 1 with pictures. Open “Show answer” for the answer and why.

How to teach this chapter · Common misconceptions & worked visual for this chapter

Combining harmonic functions and commensurate periods

Question 1

For x = 5 sin(ωt) + 12 cos(ωt) cm, what is the amplitude?

  1. (a)17 cm
  2. (b)13 cm
  3. (c)7 cm
Show answer

(b) 13 cm

The sine and cosine components are a quarter-cycle apart. Their resultant amplitude is √(a²+b²), giving 13 cm.

Energy partition, amplitude scaling and energy period

Question 2

For an unchanged ideal spring, amplitude becomes 2 times its original value. By what factor does total oscillation energy change?

  1. (a)1/2
  2. (b)4
  3. (c)2
Show answer

(b) 4

E = kspring A²/2. Energy scales as amplitude squared.

Period, frequency and angular frequency

Question 3

In the same 5 s interval, oscillator A completes 30 cycles and oscillator B completes 5 cycles. What is frequency A divided by frequency B?

  1. (a)150
  2. (b)1/6
  3. (c)6
Show answer

(c) 6

Frequency is the number of cycles per unit time, so the frequency ratio is 30/5. The period ratio is the reciprocal; treating frequency as proportional to period reverses this comparison.

Periodic, oscillatory and simple harmonic motion

Question 4

A particle executes a repeated triangular displacement-time waveform. Does repetition alone establish one-dimensional simple harmonic motion?

  1. (a)No; one-dimensional SHM additionally requires acceleration proportional and opposite to displacement from equilibrium.
  2. (b)Yes; every repeating motion is one-dimensional SHM.
  3. (c)No; one-dimensional SHM can never be periodic.
Show answer

(a) No; one-dimensional SHM additionally requires acceleration proportional and opposite to displacement from equilibrium.

Periodicity is broader than SHM. The restoring relation a = −ω²x identifies one-dimensional SHM.

Initial conditions and phase conventions

Question 5

For x = A cos(ωt+φ), A and ω are known positive constants. At t = 0, x = A/2, but the direction of motion is unknown. What can be concluded about φ in [0,2π)?

  1. (a)It cannot be determined even if initial velocity is supplied.
  2. (b)It can be π/3 or 5π/3, depending on the velocity direction.
  3. (c)It must be π/3 regardless of the velocity direction.
Show answer

(b) It can be π/3 or 5π/3, depending on the velocity direction.

The same cosine value occurs in two phase quadrants. The sign of v = −Aω sinφ selects which phase applies.

Uniform circular motion and its linear projections

Question 6

The diagram gives a uniform reference circle with the radius and period shown. At t = 0 the marked point is on the vertical axis. Using its indicated clockwise rotation, choose the x projection as a function of time t in seconds.

POxy2 mT = 4 sclockwise
  1. (a)x = −2 sin(πt/2) m
  2. (b)x = 2 cos(πt/2) m
  3. (c)x = 2 sin(πt/2) m
Show answer

(c) x = 2 sin(πt/2) m

The radius is 2 m and the period 4 s gives angular speed π/2 rad s⁻¹. For clockwise rotation, subtract angular advance from the initial angle π/2 in x = A cosθ.

Direction of velocity, acceleration and restoring force

Question 7

An SHM particle is on the positive side of equilibrium and moving farther away from equilibrium. Which direction does its acceleration have?

  1. (a)Zero acceleration until the particle reverses.
  2. (b)The negative direction.
  3. (c)The positive direction.
Show answer

(b) The negative direction.

The SHM acceleration is −ω²x. It points toward equilibrium even while the particle is moving away.

Maximum speed and speed at a given displacement

Question 8

At |x|/A = 1/2 in SHM, what is the speed divided by the maximum speed?

  1. (a)1/2
  2. (b)√3/2
  3. (c)1
Show answer

(b) √3/2

The relation is v² = ω²(A²−x²). Therefore speed/vmax = √[1−(x/A)²], not 1−|x|/A.

Pendulum equilibrium under horizontal acceleration

Question 9

A pendulum is at rest relative to a frame with constant horizontal acceleration 5 m s⁻². Gravity is 12 m s⁻². What effective gravitational magnitude sets its small-oscillation period about the tilted equilibrium?

  1. (a)13 m s⁻²
  2. (b)17 m s⁻²
  3. (c)7 m s⁻²
Show answer

(a) 13 m s⁻²

The horizontal inertial acceleration and vertical gravity are perpendicular. Their resultant magnitude is √(g²+a²), here 13 m s⁻².

Amplitude, equilibrium offset and sinusoidal displacement

Question 10

A coordinate obeys x(t) = 4 + 2 cos(ωt), in centimetres. Which description is correct?

  1. (a)It cannot be SHM because its mean coordinate is nonzero.
  2. (b)SHM about x = 4 cm, with amplitude 2 cm.
  3. (c)SHM about x = 0 with amplitude 6 cm.
Show answer

(b) SHM about x = 4 cm, with amplitude 2 cm.

Displacement from equilibrium is x minus the constant offset. That displacement obeys the standard cosine law of SHM.

Restoring linear force and the defining SHM condition

Question 11

A particle has restoring force F = −c x^5, with c > 0. Is this exact SHM for finite nonzero amplitude?

  1. (a)No; the force always points away from equilibrium.
  2. (b)No; F/x varies with displacement.
  3. (c)Yes; every force pointing toward equilibrium gives exact SHM.
Show answer

(b) No; F/x varies with displacement.

The odd power makes the force restoring. Exact SHM additionally requires F/x to be a constant negative value.

Spring period, calibration and gravity independence

Question 12

An ideal vertical mass-spring oscillator is taken from Earth to a planet with twice Earth gravity, with unchanged mass and spring stiffness. What happens to its period?

  1. (a)It changes because the local gravitational acceleration changes.
  2. (b)It stays unchanged.
  3. (c)Oscillations become impossible.
Show answer

(b) It stays unchanged.

Gravity shifts the static equilibrium of a vertical spring. Small displacements about that equilibrium still obey m xddot = −kx.

Two restoring springs and a free two-mass spring

Question 13

A mass lies between two identical horizontal springs, each stiffness 5 N m⁻¹, with fixed outer ends and no slack. It is displaced 2 cm right from equilibrium. What is the net restoring force?

  1. (a)-1/10 N
  2. (b)-1/5 N
  3. (c)0 N
Show answer

(b) -1/5 N

The left spring stretches and the right spring compresses. Both changes push or pull left, giving F = −2kx.

Small-angle pendulum and period scaling

Question 14

A pendulum is released from 60 degrees with no friction. Is its motion exactly described by the small-angle SHM equation θddot = −(g/L)θ over the whole swing?

  1. (a)No; a pendulum at large angle has no restoring torque.
  2. (b)Yes; the small-angle approximation is exact for every amplitude.
  3. (c)No; the exact restoring term is −(g/L)sinθ.
Show answer

(c) No; the exact restoring term is −(g/L)sinθ.

At a large angle, sinθ is not equal to θ in radians. The pendulum still oscillates, but the linear SHM approximation is not exact.

Buoyant cork and liquid-column restoring forces

Question 15

A uniform cylindrical cork oscillates vertically while remaining partly submerged. At fixed height and densities, its cross-sectional area is multiplied by 3. How does its small-oscillation period change?

  1. (a)It stays unchanged.
  2. (b)It is divided by √3.
  3. (c)It is multiplied by 3.
Show answer

(a) It stays unchanged.

Buoyant restoring stiffness is ρliquid gA. Cork mass is ρcork Ah, so area cancels from their ratio.

Common misconception in Oscillations

Open full misconception guide →

Where does a spring mass move fastest?

Wrong

The mass moves fastest at maximum stretch because the spring pulls hardest there.

Why it feels right. I connect the biggest force with the biggest speed.

Right

The mass changes direction at maximum displacement. In ideal spring motion its speed peaks at equilibrium.

Worked example

The mass is momentarily at rest at both ends. Its speed is greatest as it crosses equilibrium.
  1. Release a mass from maximum stretch.
  2. It starts at rest.
  3. At equilibrium, spring potential energy is at a minimum and speed is at a maximum.
  4. The mass then slows to rest at the opposite end.