Waves

Physical Pendulum Period

T=2πImgdT = 2\pi\sqrt{\frac{I}{mgd}}

Period of a rigid body oscillating about a pivot point (physical/compound pendulum).

Variables

T
Period
Time for one oscillation
Allowed values: greater than 0
s
I
Moment of inertia
About pivot point
Allowed values: greater than 0
kg·m²
m
Mass
Total mass of pendulum
Allowed values: greater than 0
kg
g
Gravitational acceleration
≈ 9.8 m/s²
Allowed values: greater than 0
m/s²
d
Distance to center of mass
From pivot to CM
Allowed values: greater than 0
m

Derivation

Torque τ = -mgd·sin(θ) ≈ -mgdθ for small angles. Using τ = Iα: Iα = -mgdθ, giving ω² = mgd/I and T = 2π√(I/mgd).

Your notes

0/5000

Constants in this formula

g
9.806 65. Exact by definition.
m/s²

Check value

Round inputs run through this page's calculator, with each constant at its tabulated value. Use it to test your own working or a rearranged form. It is a test case, not a real-world example.

  • I = 2 kg·m² (Moment of inertia)
  • m = 3 kg (Mass)
  • g = 9.806 65 m/s² (Gravitational acceleration)
  • d = 4 m (Distance to center of mass)

Result: T = 0.81911316 s

Other waves formulas with the same quantities

Where this formula is listed

Common questions

What is the Physical Pendulum Period formula?

Physical Pendulum Period is written T = 2π√(I/(mgd)). Period of a rigid body oscillating about a pivot point (physical/compound pendulum). It belongs to Waves, under Simple Harmonic Motion.

What are the units in T = 2π√(I/(mgd))?

In SI units, T is period, measured in s; I is moment of inertia, measured in kg·m²; m is mass, measured in kg; g is gravitational acceleration, measured in m/s²; d is distance to center of mass, measured in m.

How is the Physical Pendulum Period formula derived?

Torque τ = -mgd·sin(θ) ≈ -mgdθ for small angles. Using τ = Iα: Iα = -mgdθ, giving ω² = mgd/I and T = 2π√(I/mgd). The result is T = 2π√(I/(mgd)).