use a = –ω2x and recall and use, as a solution to this equation, x = x0 sin ωt

Simple Harmonic Oscillations

Simple harmonic motion (SHM) describes the oscillatory motion of a system where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction. The defining differential equation for SHM is

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

where:

• $x$ – displacement from the equilibrium position (m)
• $\omega$ – angular frequency (rad s⁻¹)
• $t$ – time (s)

Derivation of the General Solution

Integrating the second‑order differential equation yields a sinusoidal solution. One convenient form is

$$x(t) = x_{0}\sin(\omega t + \phi)$$

where:

• $x_{0}$ – amplitude (maximum displacement) (m)
• $\phi$ – phase constant (rad), determined by initial conditions

If the motion starts from the equilibrium position with maximum velocity, the phase constant $\phi$ can be set to zero, giving the simpler expression

$$x(t) = x_{0}\sin(\omega t)$$

Key Relationships

Quantity	Expression	Physical Meaning
Angular frequency	$\omega = 2\pi f = \sqrt{\dfrac{k}{m}}$	Rate of oscillation; $f$ is frequency (Hz), $k$ is spring constant (N m⁻¹), $m$ is mass (kg)
Period	$T = \dfrac{2\pi}{\omega}$	Time for one complete oscillation (s)
Velocity	$v(t) = \dfrac{dx}{dt} = \omega x_{0}\cos(\omega t)$	Instantaneous speed; maximum $v_{\max}= \omega x_{0}$
Acceleration	$a(t) = \dfrac{d^{2}x}{dt^{2}} = -\omega^{2}x_{0}\sin(\omega t) = -\omega^{2}x(t)$	Restoring acceleration; proportional to displacement
Energy	$E = \dfrac{1}{2}k x_{0}^{2} = \dfrac{1}{2}m\omega^{2}x_{0}^{2}$	Total mechanical energy (constant for an ideal SHM system)

Applying the Equation $a = -\omega^{2}x$

1. Identify the system that exhibits a linear restoring force (e.g., mass‑spring, simple pendulum for small angles).
2. Write the expression for the restoring force $F = -kx$ (Hooke’s law) or the equivalent torque for rotational systems.
3. Use Newton’s second law $F = ma$ to obtain $ma = -kx$ → $a = -(k/m)x$.
4. Recognise that $-(k/m)$ plays the role of $-\omega^{2}$, so $\omega = \sqrt{k/m}$.
5. Insert $\omega$ into the solution $x(t) = x_{0}\sin(\omega t + \phi)$ to describe the motion completely.

Example Problem

Problem: A 0.5 kg mass is attached to a horizontal spring with $k = 200\ \text{N m}^{-1}$. The mass is pulled 0.10 m from equilibrium and released from rest. Determine the displacement $x$ after $t = 0.05\ \text{s}$.

Solution:

1. Calculate the angular frequency: $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20\ \text{rad s}^{-1}$$
2. Since the mass is released from rest at maximum displacement, the phase constant $\phi = 0$ and the motion follows $$x(t) = x_{0}\sin(\omega t)$$ with $x_{0}=0.10\ \text{m}$.
3. Evaluate at $t = 0.05\ \text{s}$: $$x(0.05) = 0.10\sin(20 \times 0.05) = 0.10\sin(1.0) \approx 0.10 \times 0.8415 = 0.084\ \text{m}$$

Thus the displacement after 0.05 s is approximately $0.084\ \text{m}$.

Common Misconceptions

• Sign of acceleration: In SHM the acceleration is always opposite to the displacement, hence the negative sign in $a = -\omega^{2}x$.
• Frequency vs. angular frequency: $f$ (Hz) and $\omega$ (rad s⁻¹) are related by $\omega = 2\pi f$, not interchangeable.
• Phase constant: Ignoring $\phi$ can lead to incorrect initial conditions; always determine $\phi$ from the given start state.

Suggested Diagram