---
title: "Four collinear points of compound pendulum"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:bar-torsional-four-collinear-points"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/bar-torsional-four-collinear-points
site: BCT Engineering
---

# Four collinear points of compound pendulum

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2072 Chaitra, 5 marks; 2067 Asar, 5 marks (asked 2 times)
- Video: [Compound Pendulum: Four Co-linear Points about which Time Period is same | Part-2](https://www.youtube.com/watch?v=59Zxkj9TBHI) — 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-torsional-four-collinear-points
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

Differentiate between bar pendulum and torsional pendulum. Prove that there exist four collinear points in a bar (compound) pendulum having the same time period. Give their physical significance.

## Short answer

Bar pendulum uses gravity; torsional pendulum uses wire elasticity. From T = 2π√((k²+l²)/(gl)) the quadratic l² − (gT²/(4π²))l + k² = 0 gives two roots l₁, l₂ on each side of CG, giving four collinear points. l₁ + l₂ = L, the equivalent simple pendulum length; these are interchangeable centers of suspension and oscillation.

## Answer

#### Difference between bar pendulum and torsional pendulum
| Bar pendulum | Torsional pendulum |
|---|---|
| Swings about a horizontal axis | Twists about a vertical wire |
| Restoring torque from gravity, mgl sinθ | Restoring torque from wire's elasticity, Cθ |
| T = 2π√((k² + l²)/gl) | T = 2π√(I/C) |
| Depends on g | Independent of g |
| Used to measure g | Used to measure η and I |

#### Four collinear points
Let G be the centre of gravity, k the radius of gyration about G, and the body be suspended at a point distance *l* from G.
```text
T = 2π √((k² + l²)/(g l))
```
1. Square and rearrange: 4π² (k² + l²) = g T² l.
2. Quadratic in *l*:
```text
l² − (gT²/4π²) l + k² = 0
```
3. For a given T it has two roots l₁ and l₂ with
```text
l₁ + l₂ = gT²/4π² ,    l₁ l₂ = k²
```
So l₁ < k < l₂ (their product is k²).
4. Each root is a distance from G, and the same equation holds on either side of G. Hence there are four points with the same period: two on each side.

```text
 D        B      G      A          C
 x--------x------o------x----------x
 GD = GC = l2 ,  GB = GA = l1  (l1 < k < l2)
```
Points A and B are at distance l₁ from G, one on each side; C and D are at distance l₂ from G, one on each side. A, C lie on one side of G and B, D on the other.

5. The distance between a point at l₁ on one side and a point at l₂ on the opposite side (for example B and C, or A and D) is l₁ + l₂ = gT²/4π² = L.

#### Physical significance
- The distance l₁ + l₂ between two such non-symmetrical points on opposite sides equals the length L of the equivalent simple pendulum with the same period.
- These points are the centres of suspension and oscillation, which are interchangeable (l₂ = k²/l₁).
- Used in Kater's pendulum: when the periods about the two knife edges are equal, g = 4π² L/T² with L measured as the distance between the knife edges. This needs no knowledge of k.

## Check yourself

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

1. For a given period T, the bar pendulum equation becomes which quadratic in l?
   A. l^2 + (gT^2/4pi^2) l - k^2 = 0
   B. l^2 - (gT^2/4pi^2) l + k^2 = 0
   C. l^2 - (4pi^2/gT^2) l + k^2 = 0
   D. l^2 - gT^2 l + 4pi^2 k^2 = 0
   Answer: B. Squaring T = 2pi sqrt((k^2+l^2)/gl) gives 4pi^2(k^2+l^2) = gT^2 l. (Key point: Quadratic in l)

2. What do the two roots l1 and l2 satisfy?
   A. l1 + l2 = k^2 and l1 l2 = gT^2/4pi^2
   B. l1 = l2 = k
   C. l1 + l2 = gT^2/4pi^2 and l1 l2 = k^2
   D. l1 - l2 = gT^2/4pi^2
   Answer: C. Sum and product of roots of the quadratic give these relations, so l1 < k < l2. (Key point: Roots)

3. Why are there four points with the same period?
   A. The quadratic has four roots
   B. Because k takes two values
   C. Because T has two values
   D. Each root gives a point on both sides of G
   Answer: D. Two distances (l1, l2), each valid on either side of G, give four collinear points. (Key point: Four points)

4. What does the distance between non-symmetrical points l1 and l2 on opposite sides equal?
   A. Length of the equivalent simple pendulum
   B. The radius of gyration k
   C. The period T
   D. Twice the radius of gyration
   Answer: A. l1 + l2 = gT^2/4pi^2 = L, used in Kater's pendulum to find g. (Key point: Significance)

## Related in Engineering Physics

- [Bar vs torsional pendulum; minimum period](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Abar-torsional-similarities-minimum-period) — Engineering Physics, chapter 1
- [Torsional pendulum time period](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atorsional-pendulum-time-period) — Engineering Physics, chapter 1, asked 2×
- [Compound pendulum centres and minimum period](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acompound-pendulum-centers-minimum) — Engineering Physics, chapter 1
- [Bar pendulum minimum period condition](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Abar-pendulum-minimum-period-gyration) — Engineering Physics, chapter 1
- [Torsional pendulum, modulus of rigidity](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Abar-torsional-pendulum-rigidity) — Engineering Physics, chapter 1, asked 2×
- [Physical pendulum and interchangeability](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acompound-pendulum-interchangeability) — Engineering Physics, chapter 1, asked 2×

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