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OscillationsMedium

Simple pendulum

A simple pendulum of length L = 0.8 m is released from an angle θ₀ = 25°. Find the period (small oscillations) and the maximum speed at the lowest point.
θ₀ L
Given data
L = 0.8 mθ₀ = 25°g = 9.81 m/s²
Review the theory: Oscillazioni
Steps
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  1. For small oscillations (sinθ ≈ θ) the pendulum behaves as a harmonic oscillator. Remarkably, the period depends only on length and gravity, not on amplitude or mass (isochronism): T = 2π·√(L/g).
  2. The speed is maximum at the lowest point, where all the lost potential energy has become kinetic. Relative to the start the mass has dropped by h = L(1 − cosθ₀). From ½mv² = mgh (the mass cancels) I get v = √(2gL(1 − cosθ₀)).
Full worked solution
  1. For small oscillations (sinθ ≈ θ) the pendulum behaves as a harmonic oscillator. Remarkably, the period depends only on length and gravity, not on amplitude or mass (isochronism): T = 2π·√(L/g).
    T=2πLg=2π0.89.81T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.8}{9.81}}
    T=2π⋅0.286≈1.79 sT = 2\pi\cdot0.286 \approx \mathbf{1.79\,s} (independent of mass).
  2. The speed is maximum at the lowest point, where all the lost potential energy has become kinetic. Relative to the start the mass has dropped by h = L(1 − cosθ₀). From ½mv² = mgh (the mass cancels) I get v = √(2gL(1 − cosθ₀)).
    v=2gL(1−cos⁡θ0)v = \sqrt{2gL(1-\cos\theta_0)}
    v=2⋅9.81⋅0.8⋅0.0937=1.47≈1.21 m/sv = \sqrt{2\cdot9.81\cdot0.8\cdot0.0937} = \sqrt{1.47} \approx \mathbf{1.21\,m/s}.
Result:T≈1.79T \approx 1.79 s, vmax≈1.21v_{max} \approx 1.21 m/s
TheoryOscillazioniFormula sheetPhysics 1

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