AP Physics C: Mechanics · Unit 7: Oscillations · Lesson 7.2

Deep Dive: Frequency and Period of SHM

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
7.2.A.1Math

Period, Frequency, and Angular Frequency

Once you know a system is undergoing SHM, three related quantities describe how fast it's oscillating. The period T is the time for one full cycle. The frequency f is how many cycles happen per second. And the angular frequency ω — the same ω from x(t) = A cos(ωt + φ) — ties them both together:

T = 2π/ω = 1/f

Adjust the angular frequency ω and watch the period and frequency respond — they're three different ways of describing the exact same repeating motion.

ω (rad/s)3
one period, T
T = 2π/ω = 2.09 sf = 1/T = 0.48 Hz
💡These three quantities describe the exact same repeating motion from three different angles — knowing any one of them (ω, T, or f) lets you find the other two instantly.
7.2.A.1.iMath

Period of a Spring Oscillator

For an object attached to an ideal spring, the period has its own dedicated formula — no need to derive ω from scratch every time:

Ts = 2π√(m/k)

Adjust the mass and spring constant of an ideal mass–spring oscillator and watch its period respond — heavier masses swing more slowly, stiffer springs swing faster.

m (kg)2
k (N/m)8
preview phase
Ts = 2π√(m/k) = 3.14 s
🔑Heavier objects oscillate more slowly (larger m means larger T), and stiffer springs oscillate faster (larger k means smaller T) — and notice that amplitude doesn't appear anywhere in this formula at all.
ExampleGuided Example — Weighing an Astronaut in Zero Gravity

In orbit, an astronaut can't just step on a scale — there's no gravity to pull down on it. Instead, mission engineers strap the astronaut into a chair attached to a spring with known spring constant k = 600 N/m, and measure the period of the resulting oscillation: T = 2.1 s. Find the combined mass of the astronaut and chair.

Step 1Start from the spring-oscillator period formula
Ts = 2π√(m/k).
7.2.A.1.iiMath

Period of a Simple Pendulum

A simple pendulum displaced by a small angle has its own period formula too — and it looks strikingly different from the spring oscillator's:

Tp = 2π√(l/g)

Adjust a simple pendulum's length and try it on different worlds — notice mass never enters the formula at all.

l
l (m)1
g (m/s²)9.80
Tp = 2π√(l/g) = 2.01 s
⚠️Mass doesn't appear anywhere in Tp = 2π√(l/g) — a bowling ball and a marble on identical-length strings swing with exactly the same period. And this formula is only valid for small displacement angles; Lesson 7.5 explores what happens at larger angles and for pendulums that aren't just a point mass on a string.
ExampleGuided Example — Measuring a String's Length with a Stopwatch

You want to know the length of a long string hanging from a high ceiling, but you can't reach the top to measure it directly. You tie a small weight to the bottom, set it swinging at a small angle, and time 10 full swings: 10 periods take 25.3 s. Find the string's length. (Use g = 9.8 m/s².)

Step 1Find the period from the timing data
T = 25.3 s / 10 = 2.53 s.
← Back to Lesson 7.2Ready for 7.3? Representing and Analyzing SHM puts x(t), v(t), and a(t) on the same page.