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

# Bar vs torsional pendulum; minimum period

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2069 Chaitra, 5 marks
- Video: [lecture with notes on Compound pendulum and torsional pendulum](https://www.youtube.com/watch?v=7HeNQZoZ018) — Dr. Upendra Kumar
- 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-similarities-minimum-period
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

Point out the similarities and dissimilarities between the oscillations of a bar pendulum and a torsional pendulum. Show that the radius of gyration is equal to the distance from the center of suspension to the center of gravity of a compound pendulum, when the time period is minimum.

## Short answer

Both are angular SHM with isochronous small oscillations. Bar pendulum uses gravitational torque (T=2π√((k²+l²)/gl), torsional uses elastic torque (T=2π√(I/C)). Minimum T occurs when d(T²)/dl = 0 → l = k, so radius of gyration about CG equals distance from suspension to CG.

## Answer

#### Similarities
- In both, the restoring torque is proportional to the angular displacement, so the motion is angular SHM: I d²θ/dt² = −(constant) θ.
- Both have T = 2π√(I/restoring torque per unit angle).
- Both are isochronous: the period does not depend on amplitude (for small amplitude).
- Both exchange kinetic and potential energy with constant total energy (undamped case).

#### Dissimilarities
| Bar pendulum | Torsional pendulum |
|---|---|
| Axis horizontal; body swings in a vertical plane | Axis vertical (wire); body twists in a horizontal plane |
| Restoring torque from gravity, mgl | Restoring torque from elasticity of wire, C |
| T = 2π√((k² + l²)/gl) | T = 2π√(I/C) |
| Depends on g and sinθ ≈ θ (small angles) | Independent of g; limit is elastic limit of the wire |
| Used to find g | Used to find η and I |

#### Minimum period of a compound pendulum
With G the centre of gravity, k the radius of gyration about G and *l* the distance of the suspension point from G:
```text
T = 2π √((k² + l²)/(g l)) ,   T² = (4π²/g)(k²/l + l)
```
For minimum T: d(T²)/dl = (4π²/g)(1 − k²/l²) = 0, so l² = k², l = k.

d²(T²)/dl² = (4π²/g)(2k²/l³) > 0, so T is a minimum.

Thus when the period is minimum, k = l: the radius of gyration (about G) equals the distance from the centre of suspension to the centre of gravity, and T_min = 2π√(2k/g).

## Check yourself

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

1. Why do both the bar pendulum and the torsional pendulum perform angular SHM?
   A. Their restoring torque is proportional to the angular displacement
   B. Their restoring torque is constant in magnitude
   C. Their period depends on amplitude
   D. Both are driven by gravity
   Answer: A. Both obey I d²θ/dt² = -(constant)θ, i.e. torque proportional to angular displacement, which is the condition for angular SHM. (Key point: Restoring torque in SHM)

2. What provides the restoring torque in a bar pendulum compared with a torsional pendulum?
   A. Elasticity of the wire in both
   B. Gravity (mgl) in the bar pendulum; elasticity of the wire (C) in the torsional pendulum
   C. Gravity in the torsional pendulum; elasticity in the bar pendulum
   D. Air resistance in both
   Answer: B. The bar pendulum swings under the gravitational torque mgl, while the torsional pendulum twists against the wire's torsional constant C. (Key point: Source of restoring torque)

3. Setting d(T²)/dl = 0 for a compound pendulum, T² = (4π²/g)(k²/l + l), gives which condition?
   A. l = 2k
   B. l = k²
   C. l = k
   D. l = k/2
   Answer: C. d(T²)/dl is proportional to 1 - k²/l², which is zero when l² = k², so l = k. (Key point: Condition for minimum period)

4. When l = k, what is the minimum time period of the compound pendulum?
   A. 2π√(k/g)
   B. 2π√(2k/g)
   C. 2π√(k²/g)
   D. π√(2k/g)
   Answer: B. Substituting l = k in T = 2π√((k²+l²)/(gl)) gives T = 2π√(2k²/(gk)) = 2π√(2k/g). (Key point: Value of minimum period)

## Related in Engineering Physics

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