---
title: "Bar pendulum minimum period condition"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:bar-pendulum-minimum-period-gyration"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/bar-pendulum-minimum-period-gyration
site: BCT Engineering
---

# Bar pendulum minimum period condition

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2072 Kartik, 5 marks
- Video: [Compound Pendulum: Maximum and Minimum Time Period | Part-3](https://www.youtube.com/watch?v=cBduGvXvjQ8) — MechE Design Course
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/bar-pendulum-minimum-period-gyration
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

Show that in a bar pendulum, the minimum time period is achieved if the radius of gyration is equal to the distance of the point of suspension or point of oscillation from the center of gravity.

## Short answer

For a bar pendulum, T = 2π √((k²+l²)/(gl)). Minimizing T² gives d/dl(k²/l + l)=0 → l=k. Thus minimum period when distance from CG equals radius of gyration. Then the centre of oscillation is also at distance k, so suspension and oscillation points coincide.

## Answer

For a bar (compound) pendulum, let G be the centre of gravity, k the radius of gyration about G, and *l* the distance of the point of suspension O from G.

#### Time period
```text
I_O = Mk² + Ml²
T = 2π √(I_O/(Mgl)) = 2π √((k² + l²)/(g l))
```

#### Condition for minimum
Write T² = (4π²/g) (k²/l + l). For T to be minimum, d(T²)/dl = 0:
```text
d/dl (k²/l + l) = −k²/l² + 1 = 0
l² = k²   →   l = k
```
Second derivative: 2k²/l³ > 0 at l = k, so it is a minimum.

Hence the time period is minimum when the distance of the point of suspension from the centre of gravity equals the radius of gyration about G.

Since the centre of oscillation lies at l′ = k²/l from G on the other side, for l = k we get l′ = k²/k = k. So the distance of the point of oscillation from G is also equal to k.

```text
 O ----- G ----- O'
   <-k->   <-k->
   L = 2k
```
Minimum period: T_min = 2π √(2k/g), with equivalent simple-pendulum length L = 2k. At this point the suspension and oscillation points coincide in their distances from G, so the four collinear points merge into two.

## Check yourself

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

1. T of bar pendulum =?
   A. 2pi sqrt(l/g)
   B. 2pi sqrt(g l/(k^2 + l^2))
   C. 2pi sqrt((k^2 + l^2)/(g l))
   D. 2pi sqrt((k^2 - l^2)/gl)
   Answer: C. I_O/(Mgl). (Key point: Period)

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

3. The second derivative 2k^2/l^3 > 0 confirms?
   A. A minimum
   B. A maximum
   C. An inflection
   D. Nothing
   Answer: A. Positive curvature. (Key point: Second derivative)

4. At l = k, centre of oscillation lies at?
   A. 0 from G
   B. k from G, other side
   C. 2k from G
   D. k^2 from G
   Answer: B. l' = k^2/l = k; L = 2k. (Key point: Oscillation point)

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