AP Physics 1 · 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.1Concept

Period and Frequency

Two ways of describing the same repetition: the period, T, is how long one full oscillation takes. The frequency, f, is how many oscillations happen per second. They're reciprocals of each other — nothing new here from earlier circular-motion work:

T = 1/f

What's new in this lesson is what actually determines T for an object exhibiting SHM. That depends entirely on which physical system you're looking at — and AP Physics 1 gives you exactly two: the ideal spring oscillator, and the small-angle simple pendulum.

7.2.A.1.iMath

The Ideal Spring Oscillator's Period

For an object oscillating on an ideal spring, the period depends on exactly two things: the object's mass, and the spring's stiffness.

T = 2π√(m/k)

More mass means more inertia to overcome each cycle, so T grows with √m. A stiffer spring (larger k) pulls harder at any given displacement, speeding the oscillation up, so T shrinks as k grows. Notice there's no amplitude anywhere in this formula — how far you pull the mass back before releasing it doesn't appear at all.

ExampleWorked Example — Finding k from a Measured Period

A 0.50 kg mass hung on a spring oscillates with a measured period of 0.90 s. Find the spring constant k.

7.2.A.1.iiMath

The Simple Pendulum's Period

For a simple pendulum displaced by a small angle, the period depends on the string's length and the local strength of gravity — and, notably, not on the mass of the bob:

T = 2π√(L/g)

A longer string takes longer to swing back and forth, so T grows with √L. Stronger gravity pulls the bob back toward equilibrium harder, speeding the swing up, so T shrinks as g grows. Just like the spring formula, amplitude (how wide the swing is) doesn't appear here either — as long as the small-angle condition from Lesson 7.1 holds.

⚠️This formula is only valid for small angular displacements. It's the same boundary from 7.1.A.2.iii: once the swing angle gets too large, the restoring torque stops being proportional to angular displacement, and this exact period formula stops applying.

Adjust the sliders and watch the animation's actual rhythm change — it's running at the period computed from your formula, not just for show.

Mass m (kg)0.50
Spring constant k (N/m)20.00
Period T
0.99 s
Frequency f
1.01 Hz

Notice: dragging the amplitude of this animation (if it had one) would change nothing about T — only m and k matter.

Skill 2.DConcept

Functional Dependence — Square Roots Change Everything

Both period formulas involve a square root, and that changes how you should reason about "doubling" or "quadrupling" a variable. If a quantity inside the square root changes by a factor of n, the period changes by a factor of √n — not n itself.

ExampleGuided Example — A Tale of Two Planets

On Earth, a pendulum and a spring-mass system both have a period of 1.0 s. Both are then moved to Planet X, which has the same radius as Earth but twice the mass. Compare each system's new period to its period on Earth.

Step 1Figure out how g changes
Gravitational field strength depends on g ∝ M/R². Same radius, but mass doubles, so g_X = 2 g_Earth.
← Back to Lesson 7.2Next: Lesson 7.3 →Representing and Analyzing SHM — putting displacement, velocity, and acceleration on the same graph.