Simple Harmonic Motion

Concise answer

Any system with a linear restoring force obeys $x(t) = A\cos(\omega t + \phi)$ . Everything on the exam reduces to identifying $\omega$ for the system at hand and tracking how energy trades between kinetic and potential forms.

Definitions

Simple harmonic motion (SHM): Oscillation driven by a restoring force proportional to displacement, producing sinusoidal motion in time.
Amplitude: The maximum displacement $A$ from equilibrium during the oscillation.
Angular frequency: The rate $\omega$ at which the oscillation phase advances, in radians per second; related to the period by $\omega = 2\pi/T$.
Phase constant: The angle $\phi$ that fixes where in its cycle the oscillator starts at $t = 0$.
Damping: Energy loss to a resistive force, which causes the oscillation amplitude to decay over time.

Intuition

SHM is what happens when the pull back toward equilibrium grows in direct proportion to how far you are displaced. Doubling the amplitude doubles the restoring force at the endpoints, so the oscillator covers the larger distance in the same period.

The angular frequency is set by a stiffness-to-inertia ratio. A stiffer spring or shorter pendulum means a stronger restoring effect and faster oscillation; more mass means more inertia and slower oscillation — except for the pendulum, whose mass cancels out.

Concept walkthrough

When the net force on an object is $F = -kx$ , the motion is sinusoidal: $x(t) = A\cos(\omega t + \phi)$ with $\omega = \sqrt{k/m}$ . The period $T = 2\pi/\omega$ depends only on the stiffness and the inertia, not on the amplitude — this amplitude independence is a favorite exam fact.

A simple pendulum displaced by a small angle experiences a restoring torque approximately proportional to the angle, so it is also a simple harmonic oscillator, with $\omega = \sqrt{g/L}$ . The mass cancels, so a heavier bob swings with the same period. For large angles the small-angle approximation fails and the motion is only approximately harmonic.

In undamped SHM the total mechanical energy $\tfrac{1}{2}kA^2$ is constant: it is all potential at the turning points and all kinetic at equilibrium. With damping, a resistive force drains this energy, so the amplitude decays over time; for light damping the system still oscillates at nearly the undamped frequency, while heavy damping suppresses oscillation entirely.

After this page, you should be able to

Write the SHM position equation and read off amplitude, angular frequency, and phase.
Compute the angular frequency of a mass-spring system and of a simple pendulum.
Track the exchange between kinetic and potential energy across an oscillation cycle.
Describe qualitatively how damping changes amplitude over time.

Formulas and assumptions

SHM position equation

x(t) = A\cos(\omega t + \phi)

Variables

x: displacement from equilibrium in meters
A: amplitude in meters
omega: angular frequency in radians per second
phi: phase constant in radians

Assumptions

The restoring force is linear in displacement.
No damping or driving force acts.

Angular frequency of a mass-spring oscillator

\omega = \sqrt{\dfrac{k}{m}}

Variables

omega: angular frequency in radians per second
k: spring constant in newtons per meter
m: mass in kilograms

Assumptions

The spring is ideal and massless.
The spring obeys Hooke's law over the motion.

Angular frequency of a simple pendulum

\omega = \sqrt{\dfrac{g}{L}}

Variables

omega: angular frequency in radians per second
g: gravitational field strength in meters per second squared
L: pendulum length in meters

Assumptions

Angular displacement is small.
The bob is a point mass on a massless, inextensible string.

Period from angular frequency

T = \dfrac{2\pi}{\omega}

Variables

T: period in seconds
omega: angular frequency in radians per second

Assumptions

The motion is periodic with a single angular frequency.

Total energy in SHM

E = \tfrac{1}{2}kA^2

Variables

E: total mechanical energy in joules
k: spring constant in newtons per meter
A: amplitude in meters

Assumptions

No damping, so the total energy is constant.
Potential energy is measured from equilibrium.

Maximum speed in SHM

v_{\text{max}} = \omega A

Variables

v_max: maximum speed in meters per second, reached at equilibrium
omega: angular frequency in radians per second
A: amplitude in meters

Assumptions

No damping.
Speed is maximal where all the energy is kinetic.

Worked example

Mass-spring oscillator numbers

A $0.50\,\text{kg}$ block on a frictionless surface is attached to a spring with $k = 200\,\text{N/m}$ and released from rest at $x = 0.10\,\text{m}$ . Find the angular frequency, the period, the total energy, and the maximum speed.

1Angular frequency: $\omega = \sqrt{k/m} = \sqrt{200/0.50} = \sqrt{400} = 20\,\text{rad/s}$ .
2Period: $T = 2\pi/\omega = 2\pi/20 \approx 0.31\,\text{s}$ .
3Released from rest at $0.10\,\text{m}$ , so the amplitude is $A = 0.10\,\text{m}$ and $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(200)(0.10)^2 = 1.0\,\text{J}$ .
4Maximum speed at equilibrium: $v_{\text{max}} = \omega A = (20)(0.10) = 2.0\,\text{m/s}$ .
5Check: $\tfrac{1}{2}mv_{\text{max}}^2 = \tfrac{1}{2}(0.50)(2.0)^2 = 1.0\,\text{J}$ , matching the total energy.

$\omega = 20\,\text{rad/s}$ , $T \approx 0.31\,\text{s}$ , $E = 1.0\,\text{J}$ , and $v_{\text{max}} = 2.0\,\text{m/s}$ .

Common traps

Confusing angular frequency $\omega$ (rad/s) with frequency $f$ (Hz); they differ by a factor of $2\pi$ .
Making the pendulum period depend on the bob's mass; the mass cancels for a simple pendulum.
Assuming a larger amplitude means a longer period; in ideal SHM the period is amplitude-independent.
Placing maximum speed at the turning points instead of at equilibrium, where all the energy is kinetic.
Applying the pendulum formula at large angles where the small-angle approximation no longer holds.

GRE Physics overview Newton's Second Law and Free-Body Diagrams Energy and Momentum Conservation Sign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
University Physics Volume 1, Section 15.1: Simple Harmonic Motion — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 1, Section 15.2: Energy in Simple Harmonic Motion — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 1, Section 15.4: Pendulums — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 1, Section 15.5: Damped Oscillations — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

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Last source check: 2026-08-02
Next scheduled review: 2026-11-02

Recheck ETS content areas and the OpenStax oscillations references before each major GRE Physics preparation cycle.

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Simple Harmonic Motion

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

SHM position equation

Angular frequency of a mass-spring oscillator

Angular frequency of a simple pendulum

Period from angular frequency

Total energy in SHM

Maximum speed in SHM

Worked example

Mass-spring oscillator numbers

Common traps

Related pages and practice

Sources

Review and maintenance