Topic 7.2 Frequency and Period of SHM: relate period, frequency and angular frequency, and determine them for the mass-spring system and the simple pendulum from the system properties.

What this topic is asking

The College Board (Topic 7.2) wants you to relate period, frequency and angular frequency, and to determine them for the mass-spring system and the simple pendulum from the physical properties. These are the most-used results in the unit: the period of each standard oscillator, and how it scales with the system parameters.

Period, frequency and angular frequency

These three describe the same timing in different units, related by $\omega = 2\pi f = 2\pi/T$. The angular frequency $\omega$ is the quantity that appears directly in the differential equation and the sinusoidal solution $x = A\cos(\omega t + \phi)$, while the period and frequency are the everyday measures. Converting fluently among them, given $\omega$, find $T = 2\pi/\omega$, and so on, is essential and often the first scored step on an exam question.

The mass-spring period

For a mass $m$ on a spring of constant $k$, the angular frequency comes straight from the SHM equation $\omega^2 = k/m$, so

A stiffer spring (larger $k$) oscillates faster (shorter period); a heavier mass oscillates slower. The period grows as $\sqrt{m}$ and shrinks as $1/\sqrt{k}$, so to halve the period you quarter the mass or quadruple the stiffness. Crucially, none of these depend on the amplitude or on $g$ (the spring oscillator works the same horizontally or vertically, with gravity only shifting the equilibrium point).

The simple pendulum period

For a simple pendulum (a point mass on a light string of length $L$) at small angles, the restoring torque gives $\omega = \sqrt{g/L}$, so

The period depends on the length and the gravitational field strength but not on the bob's mass and not on the amplitude (for small swings). A longer pendulum or weaker gravity gives a longer period. This is why pendulum clocks must be calibrated for length and run differently at different altitudes or on other planets. The small-angle restriction matters: at large amplitudes the period grows slightly and the motion is no longer exactly SHM.

Try this

Q1. A mass-spring oscillator has $\omega = 10$ rad/s. Calculate its period and frequency. [2 points]

• Cue. $T = 2\pi/\omega = 0.63$ s; $f = 1/T = 1.6$ Hz.

Q2. A simple pendulum has length $1.0$ m. Calculate its period ($g = 9.8$ m/s squared). [2 points]

• Cue. $T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1.0/9.8} = 2.0$ s.