PhysicsOscillationsClass 11

Simple pendulum: interactive simulation

A pendulum is the classic example of simple harmonic motion — but only for small swings. The simulation solves the full equation θ″ = −(g/L) sin θ step by step, so you can see exactly when the textbook formula starts to drift.

Useful for: JEE Main, NEET, AP Physics 1

Try this

Pull the bob aside and let go. The θ–time graph traces a cosine, and the energy bars show kinetic and potential energy swapping every quarter swing.

  1. 1

    With g = 9.8, set L = 1 m, then L = 2 m.

    What you should see

    The period goes from about 2.01 s to 2.84 s — doubling L multiplies T by √2.

  2. 2

    Keep L fixed and raise the amplitude from 10° to 60°.

    What you should see

    At 10° the exact period is only 0.19 % longer than 2π√(L/g); at 60° it is about 7 % longer.

  3. 3

    Set g to 1.6 m/s² (the Moon).

    What you should see

    The pendulum swings about 2.5 times slower, because T ∝ 1/√g.

  4. 4

    Watch the energy bars during one swing.

    What you should see

    Kinetic energy peaks at the bottom, potential energy at the ends; their total stays constant.

Formulas behind it

Period (small angles)

T=2πLgT = 2\pi\sqrt{\frac{L}{g}}

Angular frequency

ω=gL,θ(t)=θ0cos⁡ωt\omega = \sqrt{\frac{g}{L}}, \quad \theta(t) = \theta_0\cos\omega t

Larger amplitudes (first correction)

T≈2πLg(1+θ0216)T \approx 2\pi\sqrt{\frac{L}{g}}\left(1 + \frac{\theta_0^2}{16}\right)

Energy at angle θ (per unit mass)

gL(1−cos⁡θ0)=12v2+gL(1−cos⁡θ)gL(1-\cos\theta_0) = \tfrac{1}{2}v^2 + gL(1-\cos\theta)

Common mistakes

  • Putting θ₀ in degrees into θ₀²/16 — it must be in radians.
  • Thinking a heavier bob swings faster. Mass cancels; only L and g matter.
  • Counting one swing from one side to the other as a full period. A period is there and back.
  • Using T = 2π√(L/g) for large amplitudes, where it underestimates the period.
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FAQ

What is the length of a seconds pendulum?

A seconds pendulum has T = 2 s, so L = g/π² ≈ 0.99 m on Earth (g = 9.8 m/s²).

Why is a pendulum only approximately simple harmonic?

The restoring torque depends on sin θ, not θ. For small angles sin θ ≈ θ, which gives SHM; for larger angles the period grows.

Where is the speed maximum?

At the lowest point, where all the energy is kinetic: v_max = √(2gL(1 − cos θ₀)).

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