Skip to main content

Meter stick pendulum period

Short answer

Using the parallel-axis theorem, I = mL²/12 + mh². With h = L/2 = 0.50 m, the period T = 2π√(I/mgh) = 1.64 s.

A meter stick swings about a pivot point at one end, at a distance 'h' from the stick's center of mass. Calculate the period of oscillation using the parallel axis theorem.

A uniform meter stick of mass m and length L = 1.00 m is pivoted at a point at distance h from its centre of mass (the pivot is horizontal and perpendicular to the stick).

Derivation (parallel-axis theorem)

  1. Moment of inertia about the centre of mass: I_cm = mL²/12.
  2. About the pivot, parallel-axis theorem: I = I_cm + m h² = mL²/12 + m h².
  3. Small oscillations: torque = −m g h sinθ ≈ −m g h θ, so I θ'' = −m g h θ.
  4. Angular frequency ω = √(m g h / I), so
T = 2π √(I/(m g h)) = 2π √( (L²/12 + h²) / (g h) )

Pivot at one end (h = L/2 = 0.50 m)

T = 2π √( (1/12 + 1/4) / (9.8 × 0.5) ) = 2π √(0.3333/4.9)
  = 2π √(0.06803) = 2π (0.2608)

T ≈ 1.64 s for a pivot at the end. For a general h, use T = 2π√((L²/12 + h²)/(gh)) with L = 1 m; the period is minimum at h = k = L/√12 = 0.289 m, where T = 2π√(2k/g) = 1.53 s.

Equivalent simple-pendulum length for the end pivot: 2L/3 = 0.667 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Engineering Physics chapter 1 questions →