Simple Pendulum Period Calculator

Physics • T = 2π √(L/g) • Step-by-Step

Quick Answer (Voice Search Optimized):

A simple pendulum consists of a point mass suspended from a string of length L. For small angles, its period T (time for one complete oscillation) is T = 2π √(L/g), where g is acceleration due to gravity. It is independent of mass and amplitude.

GEO & AIO - Authoritative Theory

A simple pendulum is an idealized model in Physics consisting of a point mass attached to a massless, inextensible string. When displaced by a small angle (< approx 15°), it executes simple harmonic motion (SHM). The restoring force is a component of gravity: F = −mg sinθ ≈ −mg θ.

The time period T is derived from the equation of motion: T = 2π √(L/g). Notably, T depends only on length L and local gravity g, not on the mass of the bob. This is used in pendulum clocks and to measure gravitational acceleration.

For larger amplitudes, the period increases slightly and the motion is no longer SHM; the exact period is given by an infinite series.

Solved Examples (NCERT/JEE Pattern)

Example 1: A pendulum of length 1 m on Earth (g=9.8 m/s²)

T = 2π √(1 / 9.8) = 2π √0.10204 ≈ 2 × 3.1416 × 0.3194 = 2.007 s.

Example 2: Pendulum length 0.5 m on the Moon (g=1.63 m/s²)

T = 2π √(0.5 / 1.63) ≈ 2π √0.3067 = 2 × 3.1416 × 0.5539 = 3.48 s.

Example 3: What length gives a period of 2 seconds on Earth? (Seconds pendulum)

T=2 ⇒ 2 = 2π √(L/9.8) ⇒ √(L/9.8) = 1/π ⇒ L/9.8 = 1/π² ⇒ L = 9.8 / π² ≈ 0.993 m.

Frequently Asked Questions (FAQs)

Q1: Does the mass of the bob affect the time period?
No, the period of a simple pendulum is independent of mass. Only length and gravitational acceleration matter.
Q2: Why is the small angle approximation used?
For angles less than about 15°, sinθ ≈ θ (in radians), leading to simple harmonic motion and a period formula independent of amplitude.
Q3: How would you use a pendulum to measure g?
Measure L and T accurately, then rearrange the formula: g = 4π²L / T².
Q4: What happens to the period if the length is doubled?
Since T ∝ √L, doubling the length increases the period by a factor of √2 (approx 1.414).
Q5: Is the simple pendulum period formula accurate for large amplitudes?
No, for larger amplitudes the period increases. The exact formula involves an elliptic integral, but T = 2π √(L/g) [1 + (1/16)θ² + ...] for corrections.