Topic 7.2 Frequency and Period of SHM: relate frequency and period, and calculate the period of a mass-spring system and a simple pendulum.

What this topic is asking

The College Board (Topic 7.2) wants you to relate period and frequency, and to calculate the period of the two model oscillators: a mass on a spring, $T = 2\pi\sqrt{m/k}$, and a simple pendulum, $T = 2\pi\sqrt{L/g}$. A central and heavily tested idea is that both periods are independent of amplitude.

Period and frequency

These two quantities describe how fast the oscillation repeats. A long period means a slow oscillation and a low frequency; a short period means a rapid oscillation and a high frequency. Converting between them is just taking a reciprocal, and almost every problem in the unit needs one or the other.

The mass-spring period

This formula follows from the defining condition $F = -kx$ of Topic 7.1. A larger mass resists acceleration, so it oscillates slowly; a stiffer spring provides a stronger restoring force, so it oscillates quickly. The square-root dependence is the feature most exam questions probe: you must scale the period by $\sqrt{\,}$ factors, not linearly.

The pendulum period and amplitude independence

For a simple pendulum swinging through small angles, the period is:

depending only on the length $L$ and the gravitational field strength $g$. Strikingly, it does not depend on the mass of the bob, just as free-fall does not depend on mass, because gravity provides both the restoring force and the inertia in the same proportion. A longer pendulum swings more slowly; the same pendulum on the Moon (smaller $g$) swings more slowly still. The deepest and most-tested idea in this topic is amplitude independence: for both systems, the period is set by the system's properties alone, not by how far it swings. Pull a pendulum back a little or a lot (within the small-angle range) and each full swing takes the same time; this is exactly the property that made pendulums the basis of accurate clocks. The reason is built into SHM: a larger amplitude means a larger restoring force and so a faster motion, and these two effects cancel precisely, leaving the period unchanged. Connecting the period formulas back to the linear restoring force of Topic 7.1 is the strategic key, because it explains why amplitude drops out and why mass enters the spring case but not the pendulum case.

Try this

Q1. A mass-spring system has period $T = 0.50$ s. Calculate its frequency. [1 point]

• Cue. $f = 1/T = 1/0.50 = 2.0$ Hz.

Q2. A $0.20$ kg block oscillates on a spring of constant $80$ N/m. Calculate the period. [2 points]

• Cue. $T = 2\pi\sqrt{0.20/80} = 2\pi\sqrt{0.0025} = 2\pi(0.05) = 0.31$ s.