NEET UG Physics · Chapter 11

Oscillations

Simple harmonic motion, velocity and acceleration, energy in SHM, spring and pendulum.

NCERT Class 11NEET: 1–2 Qs per paper 9 min read

1. Chapter at a Glance

A motion that repeats itself after equal intervals is periodic; if it goes back and forth about a mean position, it is an oscillation. The simplest and most important oscillation is simple harmonic motion (SHM), in which the restoring force is proportional to the displacement. This chapter covers the equation of SHM, velocity and acceleration, energy in SHM, and two standard systems — a mass on a spring and the simple pendulum. NEET asks one or two questions here almost every year.

What NEET asks most
  • Reading amplitude, frequency and phase from x = A sin(ωt + φ).
  • Maximum velocity and acceleration; velocity at a given displacement.
  • Kinetic and potential energy in SHM.
  • Period of spring systems (series and parallel) and pendulums.
  • Time to move between two positions.

2. Periodic Motion and SHM

  • Period T is the time for one complete oscillation; frequency ν = 1/T (hertz); angular frequency ω = 2πν = 2π/T.
  • Every oscillation is periodic, but not every periodic motion is an oscillation (the Earth going round the Sun is periodic but not oscillatory).
  • SHM: the restoring force F = −kx, so the acceleration a = −ω²x is proportional to displacement and directed towards the mean position.
  • Displacement: x(t) = A cos(ωt + φ) (or sine form). A is the amplitude, (ωt + φ) is the phase, φ is the phase constant.
  • SHM is the projection of uniform circular motion on a diameter.
v = ω√(A² − x²)    vmax = Aω    a = −ω²x    amax = Aω²
x = A cos ωta = −ω²xv (dashed) t
In SHM, x, v and a vary sinusoidally. Velocity is a quarter cycle out of step with displacement; acceleration is always opposite to displacement.

3. Energy in SHM

KE = ½mω²(A² − x²)    PE = ½mω²x²    E = ½mω²A² = ½kA²
  • The total energy stays constant and is proportional to A².
  • At the mean position the energy is all kinetic; at the extremes it is all potential.
  • KE = PE at x = A/√2.
  • KE and PE each vary with twice the frequency of the motion.

4. Spring and Pendulum

SystemPeriodNEET Note
Mass on a springT = 2π√(m/k)Same on the Moon (no g)
Springs in series1/k = 1/k₁ + 1/k₂Softer; longer period
Springs in parallelk = k₁ + k₂Stiffer; shorter period
Vertical spring (static stretch x₀)T = 2π√(x₀/g)Because kx₀ = mg
Simple pendulum (small angle)T = 2π√(l/g)Independent of mass and amplitude
Pendulum in a liftT = 2π√(l/(g ± a))No oscillation in free fall
  • A spring cut into n equal pieces gives each piece a spring constant nk.
  • A seconds pendulum has a period of 2 s; its length on Earth is about 1 m.
  • The simple pendulum formula holds only for small angles (sinθ ≈ θ).

5. Formula Master Table

QuantityFormula
Angular frequencyω = 2π/T = 2πν
Displacementx = A cos(ωt + φ)
Velocityv = ω√(A² − x²)
Accelerationa = −ω²x
Total energyE = ½kA² = ½mω²A²
SpringT = 2π√(m/k)
PendulumT = 2π√(l/g)
KE = PEx = A/√2

6. Master Summary Table

QuantityAt mean positionAt extreme position
Displacement0±A
SpeedMaximum (Aω)0
Acceleration0Maximum (Aω²)
Kinetic energyMaximum0
Potential energy0Maximum
Restoring force0Maximum (kA)

7. Mind Map

Oscillations
SHM
  • F = −kx
  • x = A cos(ωt + φ)
  • Phase
Kinematics
  • vmax = Aω
  • amax = Aω²
Energy
  • E = ½kA²
  • KE ↔ PE
Systems
  • Spring: √(m/k)
  • Pendulum: √(l/g)
  • Series/parallel

8. Tips and Tricks

Shortcuts
  • Time from the mean position to A/2 is T/12; from A/2 to A is T/6; mean to extreme is T/4.
  • If two speeds v₁, v₂ at positions x₁, x₂ are known, ω² = (v₁² − v₂²)/(x₂² − x₁²).
  • Pendulum length increased by p% (small p) ⇒ period increases by about p/2 %.
  • For a given x = A sin(ωt), the maximum speed is the coefficient of t times A.
Common NEET traps
  • Confusing ω with frequency (ω = 2πν).
  • Thinking a spring–mass period changes with g.
  • Saying acceleration is maximum at the mean position (it is zero there).

9. Important One-liners

  • The period of SHM does not depend on its amplitude.
  • Acceleration in SHM is always directed towards the mean position.
  • Total energy in SHM is proportional to the square of the amplitude.
  • The period of a simple pendulum does not depend on the mass of the bob.
  • A pendulum clock runs slow on a mountain because g is smaller there.
  • In a freely falling lift a pendulum stops oscillating.
  • Kinetic energy in SHM oscillates at twice the frequency of the motion.

10. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take g = 10 m s⁻² and π² ≈ 10 where convenient. Click to reveal each answer and working.

Average level

N1Average

A particle moves as x = 0.1 sin(4πt) m. Find its amplitude, frequency, period and maximum speed.

Show answer & solution

Concept: compare with x = A sin(ωt).

  1. A = 0.1 m; ω = 4π rad s⁻¹.
  2. ν = ω/2π = 2 Hz; T = 0.5 s.
  3. vmax = Aω = 0.4π ≈ 1.26 m s⁻¹.

Answer: 0.1 m; 2 Hz; 0.5 s; about 1.26 m s⁻¹

N2Average

A 2 kg block is attached to a spring of constant 200 N m⁻¹. Find ω, the period and the frequency.

Show answer & solution

Concept: ω = √(k/m).

  1. ω = √(200/2) = 10 rad s⁻¹.
  2. T = 2π/10 ≈ 0.628 s.
  3. ν = 1/T ≈ 1.59 Hz.

Answer: 10 rad s⁻¹; 0.628 s; 1.59 Hz

N3Average

A pendulum has a period of 2 s on the Earth. What will its period be on the Moon, where g is one-sixth of Earth's value?

Show answer & solution

Concept: T ∝ 1/√g.

  1. Tmoon = T√(g/gmoon) = 2√6.
  2. ≈ 4.9 s.

Answer: about 4.9 s

N4Average

A 1 kg body performs SHM with amplitude 5 cm and ω = 20 rad s⁻¹. Find its total energy.

Show answer & solution

Concept: E = ½mω²A².

  1. A = 0.05 m.
  2. E = ½ × 1 × 400 × 0.0025 = 0.5 J.

Answer: 0.5 J

N5Average

A particle oscillates with amplitude 10 cm. At what displacement are its kinetic and potential energies equal?

Show answer & solution

Concept: ½k(A² − x²) = ½kx² ⇒ x = A/√2.

  1. x = 10/√2.
  2. ≈ 7.07 cm on either side of the mean position.

Answer: about ±7.07 cm

Advanced level

N6Advanced

Springs of 300 N m⁻¹ and 600 N m⁻¹ support a 2 kg mass, first in series and then in parallel. Find the period in each case.

Show answer & solution

Concept: series 1/k = 1/k₁ + 1/k₂; parallel k = k₁ + k₂.

  1. Series: k = 300 × 600/900 = 200 N m⁻¹ ⇒ T = 2π√(2/200) ≈ 0.628 s.
  2. Parallel: k = 900 N m⁻¹ ⇒ T = 2π√(2/900) ≈ 0.296 s.

Answer: about 0.63 s; about 0.30 s

N7Advanced

The length of a simple pendulum is increased by 21%. By what percentage does its period increase?

Show answer & solution

Concept: T ∝ √l; use the exact ratio since 21% is not very small.

  1. T₂/T₁ = √1.21 = 1.1.
  2. Increase = 10%.

Answer: 10%

N8Advanced

In SHM, the acceleration is 12 cm s⁻² when the displacement is 3 cm. Find the period. If the amplitude is 5 cm, find the maximum speed.

Show answer & solution

Concept: |a| = ω²x.

  1. ω² = 12/3 = 4 ⇒ ω = 2 rad s⁻¹.
  2. T = 2π/2 = π ≈ 3.14 s.
  3. vmax = Aω = 5 × 2 = 10 cm s⁻¹.

Answer: π s; 10 cm s⁻¹

N9Advanced

A particle in SHM has period 6 s and starts from the mean position. How long does it take to go from x = A/2 to x = (√3/2)A?

Show answer & solution

Concept: x = A sin(ωt); find the phases at the two positions.

  1. x = A/2 ⇒ ωt₁ = π/6; x = (√3/2)A ⇒ ωt₂ = π/3.
  2. Phase change = π/6, which is 1/12 of a full cycle (2π).
  3. Time = T/12 = 6/12 = 0.5 s.

Answer: 0.5 s

N10Advanced

A mass hung from a vertical spring stretches it by 5 cm at equilibrium. If it is pulled down slightly and released, find the period of oscillation.

Show answer & solution

Concept: at equilibrium kx₀ = mg, so m/k = x₀/g.

  1. T = 2π√(m/k) = 2π√(x₀/g).
  2. = 2π√(0.05/10) = 2π√0.005.
  3. T ≈ 0.444 s.

Answer: about 0.44 s

11. NEET-pattern Questions with Detailed Answers

These questions are original and modelled on concepts repeatedly asked in NEET UG.

Q1Easy

In SHM, the acceleration of the particle is zero at:

  1. The extreme positions
  2. The mean position
  3. Half the amplitude
  4. All points
Show solution

a = −ω²x is zero when x = 0 (mean position), where speed is maximum.

Answer: (B)

Q2Easy

The period of a simple pendulum is independent of:

  1. Its length
  2. g
  3. The mass of the bob
  4. The planet it is on
Show solution

T = 2π√(l/g) contains neither the mass nor (for small angles) the amplitude.

Answer: (C)

Q3Medium

If the amplitude of an SHM is doubled, its total energy becomes:

  1. Double
  2. Four times
  3. Half
  4. Unchanged
Show solution

E = ½kA² ∝ A²: doubling A makes E four times. The period stays the same.

Answer: (B)

Q4Medium

A spring of constant k is cut into two equal halves. The spring constant of each half is:

  1. k/2
  2. k
  3. 2k
  4. 4k
Show solution

k ∝ 1/length; half the length gives twice the stiffness: 2k.

Answer: (C)

Q5Medium

The phase difference between the velocity and displacement of a particle in SHM is:

  1. 0
  2. π/2
  3. π
  4. 2π
Show solution

x = A sin ωt, v = Aω cos ωt = Aω sin(ωt + π/2): velocity leads by π/2. Acceleration differs from displacement by π.

Answer: (B)

Q6Medium

A pendulum in a lift has period T when the lift is at rest. When the lift accelerates upward, the period:

  1. Increases
  2. Decreases
  3. Stays the same
  4. Becomes infinite
Show solution

Effective g becomes g + a (larger), so T = 2π√(l/(g + a)) decreases.

Answer: (B)

Q7Hard

A particle in SHM has amplitude A. When its displacement is A/2, the fraction of its total energy that is kinetic is:

  1. 1/4
  2. 1/2
  3. 3/4
  4. 1
Show solution

PE/E = x²/A² = 1/4, so KE/E = 3/4.

Answer: (C)

Q8Hard

The maximum speed of a particle in SHM is 4 m s⁻¹ and its maximum acceleration is 16 m s⁻². Its amplitude is:

  1. 0.25 m
  2. 1 m
  3. 4 m
  4. 0.5 m
Show solution

amax/vmax = ω = 16/4 = 4 rad s⁻¹; A = vmax/ω = 4/4 = 1 m.

Answer: (B)

12. Quick Revision Summary

Remember before the exam
  • SHM: F = −kx, a = −ω²x, x = A cos(ωt + φ).
  • v = ω√(A² − x²); vmax = Aω; amax = Aω².
  • E = ½kA²; KE = PE at A/√2.
  • Spring T = 2π√(m/k); pendulum T = 2π√(l/g).
  • Series springs soften, parallel springs stiffen; cut spring is stiffer.
All Physics chaptersNEET Mock Tests