Mass–Spring and Pendulum Systems

Two systems appear repeatedly, and their periods are worth knowing along with the reasons behind them.

Mass on a spring —

T = 2π √(m / k)

The restoring force is F = −kx from Hooke's law, so a = −(k/m)x and comparing with a = −ω²x gives ω² = k/m. Since T = 2π/ω, the result follows.

What it depends on — a heavier mass oscillates more slowly, a stiffer spring more quickly. The amplitude does not appear, which is the isochronous property.

Example: a 0.20 kg mass on a spring of force constant 32 N m⁻¹ has T = 2π√(0.20 ÷ 32) = 2π√(0.00625) = 0.50 s.

Simple pendulum —

T = 2π √(l / g)

where l is the length from the pivot to the centre of mass of the bob.

The mass does not appear — a heavy bob and a light one on strings of equal length have the same period. The restoring force is proportional to the mass, and so is the inertia, and the two cancel — the same cancellation as in free fall.

Example: a pendulum of length 0.50 m has T = 2π√(0.50 ÷ 9.81) = 2π × 0.226 = 1.42 s.

The small-angle approximation — the pendulum formula is derived assuming sinθ ≈ θ for small angles in radians, which holds to within about 1% below roughly 10°. At larger amplitudes the period lengthens slightly and the motion is no longer strictly simple harmonic. Any experiment using this formula must keep the swing small, and saying so is worth a mark.

Measuring g with a pendulum — vary the length and time many oscillations for each, then plot T² against l. From T² = (4π²/g)l the graph is a straight line through the origin of gradient 4π²/g, so

g = 4π² ÷ gradient

Why time many oscillations — timing one swing carries the full reaction-time uncertainty on a short interval. Timing twenty and dividing spreads that same uncertainty across twenty periods, cutting the percentage uncertainty by a factor of twenty. Counting from a point of maximum speed, as the bob passes through equilibrium, is easier to judge than at the extremes, where it is momentarily stationary and the exact turning instant is hard to see.