---
title: "Compound pendulum centres and minimum period"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:compound-pendulum-centers-minimum"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/compound-pendulum-centers-minimum
site: BCT Engineering
---

# Compound pendulum centres and minimum period

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2074 Asoj, 5 marks
- Video: [Compound Pendulum | Centre of Oscillation | BSc Physics](https://www.youtube.com/watch?v=5qtveMudawU) — JIWAN THAPA PHYSICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/compound-pendulum-centers-minimum
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

Define centers of suspension and oscillation of a compound pendulum and show that they are interchangeable. What length of the pendulum has its minimum time period?

## Short answer

The centre of suspension is the pivot O; centre of oscillation is point O' on OG such that OO' = L = l + k²/l. Suspending at O' gives same period, showing interchangeability. Minimum period when distance from G equals radius of gyration k, so L=2k and T_min=2π√(2k/g).

## Answer

#### Centres of suspension and oscillation
For a compound pendulum of centre of gravity G, radius of gyration k about G and pivot O at distance *l* from G:
```text
T = 2π √((k² + l²)/(g l)) = 2π √(L/g) ,   L = l + k²/l
```
- **Centre (point) of suspension**: the pivot O about which the body oscillates.
- **Centre of oscillation**: the point O′ on the line OG, on the other side of G, at distance L from O (the length of the equivalent simple pendulum). Its distance from G is l′ = L − l = k²/l.

#### Interchangeability
Suspend at O′: l′ = k²/l, so
```text
(k² + l′²)/l′ = (k² + k⁴/l²) · l/k² = (l² + k²)/l = L
```
Hence T′ = 2π√(L/g) = T. The two points are interchangeable.

```text
 O ------ G ---- O'
   <-l->  <-k²/l->
   <------ L ----->
```

#### Minimum time period
T² = (4π²/g)(k²/l + l). Differentiate with respect to *l* and set to zero:
```text
d/dl (k²/l + l) = −k²/l² + 1 = 0   →   l = k
```
The second derivative 2k²/l³ is positive, so this is a minimum. Then
```text
L = k + k = 2k ,   T_min = 2π √(2k/g)
```
So the minimum period occurs when the distance of the pivot from G equals the radius of gyration k, i.e. the length of the equivalent simple pendulum is L = 2k.

## Check yourself

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

1. The centre of oscillation O' is at distance L from O where L is?
   A. k
   B. l - k^2/l
   C. 2l
   D. Length of the equivalent simple pendulum, l + k^2/l
   Answer: D. T = 2pi sqrt(L/g). (Key point: Centre of oscillation)

2. When suspended from O', the equivalent length is?
   A. Still L, so T' = T
   B. Zero
   C. l
   D. Larger than L
   Answer: A. (k^2+l'^2)/l' with l' = k^2/l equals L. (Key point: Interchange)

3. T is minimum when?
   A. l = 0
   B. l = k
   C. l = 2k
   D. l = k^2
   Answer: B. d/dl(k^2/l + l) = 0 gives l = k. (Key point: Minimum condition)

4. The minimum period is?
   A. 2pi sqrt(k/g)
   B. 2pi sqrt(2g/k)
   C. 2pi sqrt(2k/g)
   D. pi sqrt(k/g)
   Answer: C. L = 2k at l = k. (Key point: Minimum value)

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