Unit 7 · Topic 7.2

7.2 Frequency and Period of SHM

The period T is the time for one complete cycle, the frequency is $f = \dfrac{1}{T}$ and the angular frequency is $\omega = 2\pi f$ . A mass on a spring has $T = 2\pi\sqrt{\dfrac{m}{k}}$ , and a simple pendulum at small angles has $T = 2\pi\sqrt{\dfrac{\ell}{g}}$ . Neither depends on the amplitude.

Key terms

period
frequency
angular frequency
spring constant
amplitude

What changed in the 2024 update

Before the Fall 2024 course update, oscillations were Unit 6. They're now Unit 7, so older videos and practice may call this unit "Unit 6."

Period, frequency and angular frequency

The period T is the time for one full cycle, in seconds. The frequency f is the number of cycles per second, in hertz (Hz): f = 1/T.

The angular frequency ω tells you how fast the motion moves through its cycle, in rad/s. One cycle is 2π rad, so $\omega = 2\pi f = \frac{2\pi}{T}$ It uses the same letter as angular velocity, but here nothing has to be spinning. It's a measure of how quickly the oscillation repeats.

The amplitude A is the maximum distance from equilibrium. One full cycle means going from one side, across to the other side and back: a total distance of 4A.

Mass on a spring

For an object of mass m on an ideal spring with constant k, $\omega = \sqrt{\frac{k}{m}} \qquad T = 2\pi\sqrt{\frac{m}{k}}$ A heavier mass makes the motion slower, and a stiffer spring makes it faster.

The period doesn't depend on the amplitude. If you pull the block twice as far, it has twice as far to go, but the spring pulls twice as hard, so it moves twice as fast and the time comes out the same.

It doesn't depend on g either. A vertical spring has the same period as a horizontal one, and the same period on the Moon.

That makes a spring a way to measure mass where weighing doesn't work. Astronauts in orbit can't stand on a scale, but they can sit in a chair attached to a spring of known k, measure the period and solve for mass: $m = \dfrac{kT^2}{4\pi^2}$ .

Simple pendulum

A simple pendulum is a small mass on a light string of length ℓ. For small swings (roughly under 15°), $T = 2\pi\sqrt{\frac{\ell}{g}}$ A longer string means a longer period. Weaker gravity means a longer period.

The period doesn't depend on the mass of the bob, and for small angles it doesn't depend on the amplitude. For large swings the period gets a little longer, but you won't need the exact large-angle formula. Topic 7.5 shows where this formula comes from.

Change	Spring–mass period	Pendulum period
mass ×4	×2	no change
spring constant ×4	×½	not relevant
length ×4	not relevant	×2
g ÷ 4	no change	×2
amplitude ×2	no change	no change (small angles)

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Period of a spring–mass system
A 0.40 kg block oscillates on a spring with k = 160 N/m. Find the angular frequency, the period and the frequency.
Show the solutionHide the solution
1. Step 1: $\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{160}{0.40}} = 20$ rad/s.
2. Step 2: $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{20} \approx 0.31$ s.
3. Step 3: $f = \dfrac{1}{T} \approx 3.2$ Hz.
Answer: ω = 20 rad/s, T ≈ 0.31 s, f ≈ 3.2 Hz
Example 2Calculator allowed
Using functional dependence (classic trap)
A pendulum clock keeps perfect time on Earth. Its period on Earth is 2.0 s. It's taken to the Moon, where g is about 1/6 of Earth's. Find its period there. Would a spring–mass oscillator's period change too?
Show the solutionHide the solution
1. Step 1: $T \propto \dfrac{1}{\sqrt{g}}$ , so dividing g by 6 multiplies T by $\sqrt{6} \approx 2.45$ . The new period is about 4.9 s, so the clock runs slow.
2. Step 2: A spring–mass system has $T = 2\pi\sqrt{m/k}$ . There's no g in it, so its period is the same on the Moon.
3. Step 3: The trap is thinking heavier or lighter weight changes a spring's period. Only m and k matter.
Answer: About 4.9 s on the Moon; a spring–mass period doesn't change
Example 3Calculator allowed
Finding a length from a period
How long must a simple pendulum be to have a period of 2.0 s? Use g = 9.8 m/s².
Show the solutionHide the solution
1. Step 1: Solve $T = 2\pi\sqrt{\dfrac{\ell}{g}}$ for ℓ: $\ell = \dfrac{gT^2}{4\pi^2}$ .
2. Step 2: $\ell = \dfrac{(9.8)(2.0)^2}{4\pi^2} \approx 0.99$ m.
Answer: About 0.99 m

Common mistakes

Confusing ω (rad/s) with f (Hz). They differ by a factor of 2π.
Thinking a bigger amplitude means a longer period. For SHM, the period doesn't depend on amplitude.
Putting the bob's mass into the pendulum formula, or g into the spring formula.
Inverting the spring formula: it's √(m/k), so more mass means a longer period.

On the exam

Ratio questions are very common: if m doubles, T grows by √2. Work with the proportionality instead of recalculating from scratch.
Lab questions may ask you to find k or g from data. Plot T² against m (spring) or against ℓ (pendulum) to get a straight line, and use the slope.

Connected topics

Videos

Topic 7.2 - Frequency and Period of SHM
Lessons With LondotWatch on YouTube (opens in a new tab)
Simple Harmonic Motion Derivations using Calculus (Mass-Spring System)
Flipping PhysicsWatch on YouTube (opens in a new tab)
Horizontal Mass–Spring System Explained | AP Physics 1 - Unit 7 Lesson 3
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Period dependence for mass on spring | Physics | Khan Academy
Khan Academy PhysicsWatch on YouTube (opens in a new tab)
AP Physics C - Springs
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
How To Solve Simple Harmonic Motion Problems In Physics
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 7.2 Frequency and Period of SHM. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A 0.80 kg block on a horizontal spring with spring constant 20 N/m oscillates on a frictionless surface. What is its period?

Question 2 of 4Calculator allowed

A block oscillates on a spring with period T. It is replaced by a block with 4 times the mass. What is the new period?

Question 3 of 4Calculator allowed

A simple pendulum is 2.5 m long and swings through small angles. Use g = 10 m/s². What is its period?

Question 4 of 4Calculator allowed

A pendulum clock is taken to the Moon, where the gravitational field is about one-sixth of Earth's. By about what factor does its small-angle period change?

0 of 4 answered