Simple Pendulum Lab: What Sets the Swing Time?

Start simple: a swing keeps its own time

Sit on a playground swing and give one push. You glide forward, slow, pause, and come back, over and over. Here is the surprising part: whether you swing in a big arc or a small one, each round trip takes almost the same amount of time. A heavier friend on the same swing keeps the same rhythm too. The only thing that really changes the timing is the length of the ropes: a long swing is slow and lazy, a short one is quick and busy. That steady back-and-forth beat is what a pendulum is all about.

Build it up: what sets the timing?

One full back-and-forth is called the period, T. Two things control it: the length L of the string and the strength of gravity g pulling the bob down. Longer string → longer period; stronger gravity → shorter period. For gentle swings these combine into one clean formula:

Period of a simple pendulum (small swing): T = 2π√(L/g)

Try a real number: a 1 m string on Earth (g = 9.81 m/s²) gives T = 2π√(1/9.81) ≈ 2.0 s, about one second to swing out and one to swing back. Notice what is missing: the mass m and the size of the swing do not appear at all.

Why, and where it breaks (AP / intro college)

Gravity pulls straight down, but only the part along the arc, −mg sin θ, pushes the bob back toward the bottom. Newton’s law along the arc gives d²θ/dt² = −(g/L) sin θ. For small angles sin θ ≈ θ, so the restoring pull is proportional to the displacement (the signature of simple harmonic motion), and out pops T = 2π√(L/g). Because m cancels (gravity pulls harder on a heavy bob but inertia resists exactly as much), mass drops out. At large angles sin θ < θ, so the restoring pull is a little weaker than SHM assumes and every swing takes longer: about 1.7 % longer at 30° and 18.0 % longer at 90°.

Equation derivation

Governing equation, nonlinear damped pendulum:

d²θ/dt² + (b/m) dθ/dt + (g/L) sin θ = 0

Small-angle limit (θ₀ ≲ 15°, sin θ ≈ θ; the period is off by under 0.5 %):

d²θ/dt² + (g/L) θ = 0  ⇒  T = 2π√(L/g),   f = (1/2π)√(g/L)

1. Tangential restoring force. Of the bob’s weight mg, only the part along the arc, F_t = −mg sin θ, pulls it back. The string tension balances the part along the string.
2. Newton’s second law. Arc length s = Lθ, so the acceleration along the arc is a_t = L d²θ/dt². Then mL d²θ/dt² = −mg sin θ, so d²θ/dt² = −(g/L) sin θ.
3. String tension. Along the string: T_s = m(g cos θ + Lω²). It is largest at the bottom of the swing (θ = 0), where the bob is fastest.
4. Linear damping. Air resistance adds −(b/m) dθ/dt. The damped angular frequency is ω_d = √(ω₀² − (b/2m)²), so the period gets slightly longer.
5. Small-angle simple harmonic motion. For θ₀ ≲ 15°: θ(t) = θ₀ e^−bt/2m cos(ω_dt + φ). The period T = 2π√(L/g) does not depend on the mass or the amplitude.
6. Large-angle correction. The exact period is T = 2π√(L/g) (1 + θ₀²/16 + 11θ₀⁴/3072 + …), always longer than the small-angle value (exact form: an elliptic integral, see the table above).
7. Numerical method. The simulators use 4th-order Runge–Kutta (RK4) with small time steps (a few milliseconds or less), accurate at all amplitudes and in every mode, including the chaotic double pendulum.

Energy. At the ends KE = 0 and PE = mgh with h = L(1 − cos θ); at the bottom all of it is kinetic, so v_max = √(2gL(1 − cos θ₀)): 0.55 m/s for a 1 m pendulum let go at 10°. A graph of T² against L is a straight line with slope 4π²/g = 4.02 s²/m on Earth, a classic way to measure g.

High school misconceptions

References: OpenStax College Physics 2e §16.4; Halliday, Resnick & Walker, Fundamentals of Physics, 10th ed., §15-4 and §15-6; Serway & Jewett, Physics for Scientists and Engineers, 8th ed., §15.5; Kleppner & Kolenkow, An Introduction to Mechanics, 2nd ed., §7.4; Marion & Thornton, Classical Dynamics, §4.3; Strogatz, Nonlinear Dynamics and Chaos, ch. 9; Arons, A Guide to Introductory Physics Teaching (1990).