Simple Harmonic Motion

Branch note: This page deepens one part of Oscillations and Simple Harmonic Motion.

Overview

Simple harmonic motion (SHM) is the most important ideal model of oscillatory motion. It describes motion in which the restoring effect causes acceleration toward equilibrium, with magnitude proportional to displacement from equilibrium.

Many physical systems approximate SHM for small displacements, including spring-mass systems, simple pendulums at small angles, vibrating molecules, tuning forks, and electrical oscillators.

Core Ideas

• SHM is defined by acceleration proportional to displacement and directed toward equilibrium.
• The vector form $\vec{a}=-{\omega }^2\vec{s}$ is the strict general statement.
• In one-dimensional problems, $a=-{\omega }^{2}x$ is the signed-component form after choosing a positive direction.
• Ideal SHM has sinusoidal displacement, velocity, and acceleration.
• Velocity leads displacement by $\pi /2$, while acceleration is in antiphase with displacement.
• Total mechanical energy is constant in ideal SHM, transferring between kinetic and potential forms.

Exam Relevance

SHM questions usually test whether a system really satisfies the restoring-acceleration condition, whether graph phase relationships are interpreted correctly, and whether energy or velocity-displacement relations are chosen appropriately.

You should be able to:

• state the SHM condition in vector and one-dimensional signed forms
• derive $a=-{\omega }^{2}x$ from a sinusoidal displacement expression
• use $v_{\max }=\omega x_0$ and $a_{\max }={\omega }^2{x}_0$
• interpret $x$-$t$, $v$-$t$, $a$-$t$, $a$-$x$, and $v$-$x$ graphs
• apply the energy relations for ideal SHM

Definition

A particle executes simple harmonic motion when its acceleration is:

• directly proportional to its displacement from equilibrium;
• always directed toward equilibrium.

Vector form:

$$\vec{a}=-{\omega }^2\vec{s}$$

In one-dimensional signed-component form, after choosing a positive direction:

$$a=-{\omega }^{2}x$$

where $x$ is the signed displacement component from equilibrium, $a$ is the signed acceleration component, and $\omega$ is the scalar angular frequency.

Why It Matters

SHM is fundamental because it produces sinusoidal motion and forms the basis of wave theory, resonance, and many advanced topics in physics. It also gives a clean way to connect force, motion, phase, and energy.

The central modelling test is whether the restoring acceleration or restoring force is proportional to displacement and directed back toward equilibrium.

Key Representations

Why SHM Occurs

SHM arises when the restoring force is proportional to displacement. If:

$$\vec{F}=-k\vec{s}$$

then by Newton’s second law:

$$m\vec{a}=-k\vec{s}$$

so:

$$\vec{a}=-\frac{k}{m}\vec{s}$$

Comparing with $\vec{a}=-{\omega }^2\vec{s}$ gives:

$$\omega =\sqrt{\frac{k}{m}}$$

This is why ideal springs naturally produce SHM.

In one-dimensional signed-component form, the same logic is often written more explicitly as:

$$F=-kx$$

Then:

$$ma=-kx$$

so:

$$a=-\frac{k}{m}x$$

Comparing with:

$$a=-{\omega }^{2}x$$

shows that:

$${\omega }^2=\frac{k}{m}$$

The negative sign gives the restoring direction. The proportionality to $x$ gives the harmonic relation.

Displacement, Velocity, and Acceleration

The displacement can be written as:

$$x=x_0\sin (\omega t+{\varphi }_0)$$

or:

$$x=x_0\cos (\omega t+{\varphi }_0)$$

where $x_0$ is the positive amplitude and ${\varphi }_0$ is the initial phase constant. The instantaneous phase is $\Phi (t)=\omega t+{\varphi }_0$.

Figure: The phase constant selects the initial state. Both the initial displacement and the initial slope are needed: the slope gives the sign of the initial velocity.

If:

$$x=x_0\sin (\omega t+{\varphi }_0)$$

then:

$$v=\frac{dx}{dt}=\omega x_0\cos (\omega t+{\varphi }_0)$$

Here $v$ is the signed velocity component. The speed is $|v|$.

Maximum speed:

$$v_{\max }=\omega x_0$$

Acceleration is:

$$a=\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$$

Maximum acceleration magnitude:

$$a_{\max }={\omega }^2{x}_0$$

This chain shows how the standard SHM equations are related mathematically:

1. start with a sinusoidal displacement function
2. differentiate once to get velocity
3. differentiate again to get acceleration
4. compare the final result with the original displacement

For

$$x=x_0\sin (\omega t+{\varphi }_0)$$

we obtain:

$$v=\omega x_0\cos (\omega t+{\varphi }_0)$$

and:

$$a=-{\omega }^2{x}_0\sin (\omega t+{\varphi }_0)$$

Since $x=x_0\sin (\omega t+{\varphi }_0)$, it follows that:

$$a=-{\omega }^{2}x$$

So the sinusoidal form and the SHM condition are two consistent descriptions of the same motion.

Time Quantities

Period:

$$T=\frac{2\pi }{\omega }$$

Frequency:

$$f=\frac{1}{T}$$

Therefore:

$$\omega =2\pi f$$

Graphical Behaviour

The displacement-time graph is sinusoidal. The velocity-time graph is also sinusoidal and shifted by $\pi /2$ relative to displacement. The acceleration-time graph is sinusoidal and in antiphase with displacement.

The acceleration-displacement graph is a straight line:

$$a=-{\omega }^{2}x$$

with gradient:

$$-{\omega }^2$$

This is a common test for SHM.

Figure: The vertically aligned graphs share the same time markers. At an extreme, $v=0$ and $|a|$ is maximum; at equilibrium, $|v|$ is maximum and $a=0$.

Figure: A straight $a$-$x$ line through the origin with negative gradient is the graphical signature of SHM; $\omega =\sqrt{-\text{gradient}}$.

Phase Relationships

If:

$$x=x_0\sin \omega t$$

then:

$$v=\omega x_0\cos \omega t$$

Velocity leads displacement by:

$$\frac{\pi }{2}$$

Acceleration is:

$$a=-{\omega }^{2}x$$

so acceleration is in antiphase with displacement. See Phase Difference.

Useful trigonometric identities behind these phase relationships are: $\sin (\theta +\pi /2)=\cos \theta$, $\cos (\theta +\pi /2)=-\sin \theta$, and ${\sin }^2\theta +{\cos }^2\theta =1$.¹

Velocity-Displacement Relation

Eliminating time:

$$v^2={\omega }^2(x_0^2-x^2)$$

Hence:

$$v=\pm \omega \sqrt{x_0^2-x^2}$$

This is useful when time is not given. The magnitude is fixed by position, but the sign is not: use the stated direction of motion, the relevant branch of an $x$-$v$ graph, or the initial condition to select $+$ or $-$.

Figure: The upper branch has $v>0$ and the lower branch has $v<0$. The two branches show why the same displacement can occur while the oscillator moves in either direction.

Energy in SHM

For ideal SHM, total mechanical energy remains constant:

$$E_T=E_K+E_P$$

Kinetic energy:

$$E_K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2(x_0^2-x^2)$$

Potential energy associated with the restoring force, choosing zero at equilibrium:

$$E_P=\frac{1}{2}m{\omega }^2{x}^2$$

Total energy:

$$E_T=\frac{1}{2}m{\omega }^2{x}_0^2$$

Speed and kinetic energy are maximum at equilibrium. Acceleration magnitude and potential energy are maximum at turning points.

Figure: Against displacement, $E_P/E_T=(x/x_0{)}^2$ and $E_K/E_T=1-(x/x_0{)}^2$. At every position their sum is the horizontal total-energy line.

Figure: $E_K$ and $E_P$ each repeat after $T/2$, because changing the sign of $x$ or $v$ does not change a squared energy term. Total energy remains constant only for the ideal undamped model.

Physical Examples

For a spring-mass system:

$$T=2\pi \sqrt{\frac{m}{k}}$$

For a simple pendulum at small angle:

$$T=2\pi \sqrt{\frac{l}{g}}$$

See Pendulum Motion for the pendulum assumptions and derivation.

For a vertical spring-mass system, weight shifts the equilibrium position, but oscillation about that new equilibrium still obeys:

$$F=-ky$$

where $y$ is displacement from the equilibrium position, so the motion is still SHM with the same:

$$T=2\pi \sqrt{\frac{m}{k}}$$

For a floating object with uniform cross-sectional area $A$ in a liquid of density $\rho$, a small vertical displacement $x$ changes the upthrust by:

$$\Delta U=\rho gAx$$

so the restoring force has the form:

$$F=-\rho gAx$$

which also implies SHM for small vertical oscillations about equilibrium.

Enrichment — Circular Motion Interpretation

Enrichment — beyond the explicit 9749 Oscillations outcomes

This projection is a useful visual derivation of the sinusoidal solution, but it is not a separate formula or path that an SHM oscillator must follow.

SHM can be viewed as the projection of uniform circular motion onto one diameter. If a particle moves in a circle of radius $r$ with angular velocity vector $\vec{\omega }$, then $\omega =|\vec{\omega }|$ and projected motion is:

$$x=r\sin \omega t$$

Thus amplitude corresponds to radius, and angular frequency is the same as the circular motion.

Common Mistakes

• Forgetting the negative sign in $a=-{\omega }^{2}x$.
• Confusing amplitude with instantaneous displacement.
• Assuming all oscillations are SHM.
• Forgetting velocity is zero at turning points.
• Using the pendulum SHM formula at large angles.

Summary

SHM is the special oscillation where acceleration is proportional to displacement and directed toward equilibrium. The defining equation $a=-{\omega }^{2}x$ links the motion, while the sinusoidal displacement, velocity, and acceleration graphs show the fixed phase relationships that appear repeatedly in exam questions.

Links

• Main topic: Oscillations and Simple Harmonic Motion
• Related concept: Phase Difference
• Related branch: Spring Motion
• Related branch: Floating Object SHM
• Related branch: Pendulum Motion
• Related branch: Damping and Resonance
• Related topic: Circular Motion

Footnotes

1. The first two identities explain why a phase shift of $\pi /2$ turns a sine form into the corresponding cosine form, and vice versa. The identity ${\sin }^2\theta +{\cos }^2\theta =1$ is what leads to the standard SHM ellipse in the $x$-$v$ plane when $x/x_0=\sin (\omega t+{\varphi }_0)$ and $v/(\omega x_0)=\cos (\omega t+{\varphi }_0)$. ↩