---
title: "Compound pendulum with added mass"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:compound-pendulum-added-mass"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/compound-pendulum-added-mass
site: BCT Engineering
---

# Compound pendulum with added mass

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2069 Asar, 5 marks
- 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-added-mass
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

Obtain an expression for the time period of a compound pendulum and show that its time period is unaffected by the fixing of a small additional mass to it at its center of suspension.

## Short answer

Compound pendulum period T = 2π√((k² + l²)/gl). A small mass fixed at the suspension point adds zero moment of inertia about the axis and its weight produces zero torque because the lever arm is zero, so the period remains unchanged.

## Answer

#### Time period of a compound pendulum
Let M be the mass, O the point of suspension, G the centre of gravity at distance *l* from O, and k the radius of gyration about G.

1. Small angular displacement θ: restoring torque = −Mgl sinθ ≈ −Mglθ.
2. Moment of inertia about O: I = Mk² + Ml².
3. I d²θ/dt² = −Mgl θ gives d²θ/dt² + [Mgl/I] θ = 0, SHM with ω² = Mgl/I.
```text
T = 2π √(I/(Mgl)) = 2π √((k² + l²)/(g l))
```

#### Effect of a small mass fixed at the point of suspension
Fix a small mass m at O itself.

1. Moment of inertia of m about the axis through O: m × (distance from O)² = m × 0² = 0. So the total moment of inertia is unchanged: I′ = I.
2. Torque of its weight about O: mg × 0 = 0, because the lever arm is zero. So the total restoring torque is still Mglθ.
3. Both the moment of inertia about the axis and the restoring torque are unchanged, hence
```text
T′ = 2π √(I/(Mgl)) = T
```
The time period is unaffected by the small mass fixed at the centre of suspension.

## Check yourself

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

1. T of compound 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/g)
   Answer: C. I/(Mgl). (Key point: Period)

2. Small mass m at O adds to I about O?
   A. m l^2
   B. m k^2
   C. m g
   D. Zero, since distance is zero
   Answer: D. I = m r^2 with r = 0. (Key point: Moment of inertia)

3. Torque of its weight about O?
   A. Zero, lever arm is zero
   B. mg l
   C. mg k
   D. mg
   Answer: A. Force through the pivot. (Key point: Torque)

4. So the period?
   A. Increases
   B. Unchanged
   C. Decreases
   D. Becomes zero
   Answer: B. Neither I nor torque changes. (Key point: Conclusion)

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