Pendulums: Simple, Physical, and Spring

Introduction

Pendulums exhibit SHM for small oscillations. Simple pendulum (point mass on string), physical pendulum (extended body), and spring-mass systems all follow similar mathematics with different effective parameters.

Simple Pendulum

For mass m on length L string: Restoring torque τ = -mgLsinθ ≈ -mgLθ for small θ. Equation: \(\ddot{\theta}\) + (g/L)θ = 0. Angular frequency ω = √(g/L). Period T = 2π√(L/g). Independent of mass, depends only on length and g.

Physical Pendulum

Extended body pivoting about point O at distance d from CM. Torque τ = -mgdsinθ. Moment of inertia I about pivot. Equation: \(\ddot{\theta}\) + (mgd/I)θ = 0. ω = √(mgd/I). Equivalent length L_eq = I/(md).

Radius of Gyration

I = mk² where k is radius of gyration. Then ω = √(gd/k²). For compound pendulum, define equivalent simple pendulum length L = k²/d. Period T = 2π√(L/g) same form as simple pendulum.

Spring-Mass System

Mass m on spring k: ω = √(k/m). Period T = 2π√(m/k). Vertical spring: same ω, equilibrium position shifted. Two springs: series 1/k_eq = 1/k₁ + 1/k₂, parallel k_eq = k₁ + k₂.

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Introduction

Pendulums exhibit SHM for small oscillations. Simple pendulum (point mass on string), physical pendulum (extended body), and spring-mass systems all follow similar mathematics with different effective parameters.

Simple Pendulum

For mass m on length L string: Restoring torque τ = -mgLsinθ ≈ -mgLθ for small θ. Equation: \(\ddot{\theta}\) + (g/L)θ = 0. Angular frequency ω = √(g/L). Period T = 2π√(L/g). Independent of mass, depends only on length and g.

Physical Pendulum

Extended body pivoting about point O at distance d from CM. Torque τ = -mgdsinθ. Moment of inertia I about pivot. Equation: \(\ddot{\theta}\) + (mgd/I)θ = 0. ω = √(mgd/I). Equivalent length L_eq = I/(md).

Radius of Gyration

I = mk² where k is radius of gyration. Then ω = √(gd/k²). For compound pendulum, define equivalent simple pendulum length L = k²/d. Period T = 2π√(L/g) same form as simple pendulum.

Spring-Mass System

Mass m on spring k: ω = √(k/m). Period T = 2π√(m/k). Vertical spring: same ω, equilibrium position shifted. Two springs: series 1/k_eq = 1/k₁ + 1/k₂, parallel k_eq = k₁ + k₂.

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Torsional Pendulum

Disk suspended by wire: restoring torque τ = -κθ where κ is torsional constant. I\(\ddot{\theta}\) = -κθ. ω = √(κ/I). Used in Cavendish experiment to measure G. Period depends on wire stiffness and disk moment of inertia.

Solved Example: Clock Pendulum

A grandfather clock needs T = 2 s (1 s each half-swing). (a) Find required simple pendulum length. (b) If clock runs slow (T = 2.1 s), how to adjust? Solution: (a) T = 2π√(L/g) → T² = 4π²L/g → L = gT²/(4π²) = 9.8 × 4 / (4π²) = 39.2/39.48 = 0.993 m ≈ 1.0 m. (b) Clock runs slow means T is too long, so L is too long. Shorten pendulum. New L needed: L' = g(2.1)²/(4π²) = 9.8 × 4.41/39.48 = 1.09 m. Current L is about 1.09 m but should be 0.99 m. Shorten by about 10 cm. Alternatively, fractional change: ΔT/T = 1/2 ΔL/L → 0.1/2 = 1/2 ΔL/L → ΔL/L = 0.1 → 10% reduction needed. For 1 m pendulum, shorten by 10 cm.

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