Q1. Find the length of a pendulum whose time period is 2 s (a "seconds pendulum", g = 9.8).
T = 2π√(L/g) ⇒ L = gT²/(4π²)
L = 9.8 × 4 / (4 × 9.87)
L ≈ 0.993 m ≈ 1 m
Oscillations — Class 11
Pendulums, springs and even atoms oscillate. Learn simple harmonic motion and compute the period of a pendulum or a spring.
Simple harmonic motion (SHM)Oscillation where the restoring force is proportional to displacement and directed towards the mean position: a = −ω²x.
Simple pendulumFor small angles, time period T = 2π√(L/g); it depends on length and g, not on mass or amplitude.
Spring–mass systemA mass m on a spring of constant k oscillates with T = 2π√(m/k).
Energy in SHMTotal energy = ½mω²A² stays constant, continually swapping between kinetic and potential.
Oscillation period calculator Interactive
Find the time period of a simple pendulum and of a spring–mass system (g = 9.8 m/s²).
Pendulum, 2π√(L/g)—s
Spring, 2π√(m/k)—s
A pendulum's period does not depend on its mass or (small) amplitude — only on length and g.
Numericals practice Interactive
Pendulum and spring problems with steps.
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Solved numericals
Q2. A mass of 2 kg on a spring of constant 200 N/m is displaced and released. Find its frequency.
T = 2π√(m/k) = 2π√(2/200) = 2π × 0.1 = 0.628 s
f = 1/T = 1/0.628
f ≈ 1.59 Hz
Formula sheet
| Quantity | Formula | SI unit |
|---|---|---|
| Time period (pendulum) | T = 2π √(L/g) | second (s) |
| Time period (spring) | T = 2π √(m/k) | s |
| SHM acceleration | a = −ω² x | m/s² |
| Total energy in SHM | E = ½ m ω² A² | J |
Common mistakes & exam wins
- A pendulum's period depends only on length and g — not on the bob's mass or (small) amplitude.
- SHM has acceleration a = −ω²x — always directed towards the mean position.
- Total energy in SHM is constant, swapping between kinetic (at the centre) and potential (at the ends).
- Angular frequency ω = 2π/T = 2πf.
Frequently asked questions
What is simple harmonic motion?
Oscillatory motion in which the restoring force (and acceleration) is proportional to the displacement and directed towards the mean position: a = −ω²x.
What does the period of a simple pendulum depend on?
Only on its length and the acceleration due to gravity, T = 2π√(L/g) — not on the mass or (small) amplitude.
How does the period of a spring–mass system depend on mass?
It increases with the square root of the mass: T = 2π√(m/k).