AcademyOscillations
Academy
Simple Harmonic Motion
Level 1 - Physics topic page in Oscillations.
Principle
Simple harmonic motion has acceleration opposite and proportional to displacement.
Notation
\(x\)
displacement from equilibrium
\(\mathrm{m}\)
\(v\)
velocity
\(\mathrm{m\,s^{-1}}\)
\(a\)
acceleration
\(\mathrm{m\,s^{-2}}\)
\(\omega\)
angular frequency
\(\mathrm{rad\,s^{-1}}\)
\(k\)
spring constant
\(\mathrm{N\,m^{-1}}\)
\(m\)
oscillating mass
\(\mathrm{kg}\)
\(A\)
amplitude
\(\mathrm{m}\)
\(\phi\)
phase constant
\(\mathrm{rad}\)
Method
Derivation 1: Start from a restoring force
A restoring force points back toward equilibrium. For a spring-like oscillator, that force is linear in displacement.
Linear restoring force
\[F=-kx\]
Apply Newton's law
\[m\frac{d^2x}{dt^2}=-kx\]
Isolate acceleration
\[\frac{d^2x}{dt^2}=-\frac{k}{m}x\]
Define angular frequency
\[\omega^2=\frac{k}{m}\]
This defines \(\omega\) as the characteristic frequency of the system.
Standard SHM equation
\[\frac{d^2x}{dt^2}=-\omega^2x\]
Derivation 2: Solve the differential equation
The general solution to \(\frac{d^2x}{dt^2}=-\\omega^2x\) is a linear combination of sine and cosine.
Try exponential form
\[x=e^{\lambda t}\Rightarrow\lambda^2e^{\lambda t}=-\omega^2e^{\lambda t}\]
Characteristic equation
\[\lambda^2=-\omega^2\Rightarrow\lambda=\pm i\omega\]
Complex roots mean the solution is oscillatory, not exponential.
General solution
\[x(t)=C_1\cos\omega t+C_2\sin\omega t\]
Any linear combination of sine and cosine is also a solution.
Combine using phase
\[C_1\cos\omega t+C_2\sin\omega t=A\cos(\omega t+\phi)\]
The amplitude A and phase constant \(\phi\) are determined by initial conditions.
Find A and \(\phi\)
\[A=\sqrt{C_1^2+C_2^2},\qquad\tan\phi=\frac{C_2}{C_1}\]
Derivation 3: Verify the sinusoidal solution
Differentiating twice reproduces the original function with a negative factor, confirming the sinusoidal form.
Sinusoidal position
\[x=A\cos(\omega t+\phi)\]
First derivative (velocity)
\[v=\frac{dx}{dt}=-A\omega\sin(\omega t+\phi)\]
Second derivative (acceleration)
\[a=\frac{d^2x}{dt^2}=-A\omega^2\cos(\omega t+\phi)\]
Recover displacement
\[a=-\omega^2x\]
This matches the SHM condition exactly.
Derivation 4: Convert angular frequency to period
One full oscillation is one phase increase of \(2\\pi\).
One cycle
\[\omega T=2\pi\]
Period
\[T=\frac{2\pi}{\omega}\]
Spring period
\[T=2\pi\sqrt{\frac{m}{k}}\]
Rules
These are the compact results from the derivations above.
SHM condition
\[a=-\omega^2x\]
Angular frequency
\[\omega=\sqrt{\frac{k}{m}}\]
Position solution
\[x=A\cos(\omega t+\phi)\]
Velocity solution
\[v=-A\omega\sin(\omega t+\phi)\]
Oscillation period
\[T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{k}}\]
Examples
Question
A mass \(m\) on a spring \(k\) starts from rest at
\[x=A\]
Write \[x(t)\]
Answer
Starting from rest at maximum displacement means
\[v(0)=0\]
and \[x(0)=A\]
With \[x(t)=A\cos(\omega t+\phi)\]
we need \[\cos\phi=1\]
so \[\phi=0\]
The result is \[x(t)=A\cos\left(\sqrt{\frac{k}{m}}t\right)\]
Checks
- Acceleration always points toward equilibrium.
- Larger mass lowers the oscillation frequency.
- Larger spring constant raises the oscillation frequency.
- The phase constant \(\\phi\) is determined by initial conditions, not the system properties.
- SHM requires a linear restoring force model.