---
title: "Meter stick pendulum period"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:meter-stick-pendulum-period"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/meter-stick-pendulum-period
site: BCT Engineering
---

# Meter stick pendulum period

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2067 Asoj, 5 marks
- Video: [Physical pendulum period derived and rigid rod period of a physical pendulum example.](https://www.youtube.com/watch?v=qmpQV10SUaI) — Zak's Lab
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/meter-stick-pendulum-period
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

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.

## 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.

## Answer

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
```text
T = 2π √(I/(m g h)) = 2π √( (L²/12 + h²) / (g h) )
```

#### Pivot at one end (h = L/2 = 0.50 m)
```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Using the parallel-axis theorem, the moment of inertia of the stick about a pivot at distance h from its centre is
   A. mL²/12 - mh²
   B. mL²/3
   C. mh²
   D. mL²/12 + mh²
   Answer: D. I = I_cm + mh² with I_cm = mL²/12. (Key point: Moment of inertia about pivot)

2. The period for small oscillations is
   A. 2π√((L²/12 + h²)/(gh))
   B. 2π√(gh/(L²/12 + h²))
   C. 2π√(h/g)
   D. 2π√((L²/12 - h²)/(gh))
   Answer: A. T = 2π√(I/(mgh)) with I = m(L²/12 + h²). (Key point: Period expression)

3. For a 1.00 m stick pivoted at its end (h = 0.5 m), the period is about
   A. 1.00 s
   B. 2.05 s
   C. 1.64 s
   D. 0.82 s
   Answer: C. T = 2π√(0.3333/4.9) = 2π(0.2608) ≈ 1.64 s. (Key point: End-pivot period)

4. At what h is the period of the stick minimum?
   A. h = L/2
   B. h = L/√12 (the radius of gyration)
   C. h = L/12
   D. h = L
   Answer: B. The period is least when h = k = L/√12 = 0.289 m, giving T ≈ 1.53 s. (Key point: Minimum period)

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