$\frac{d^2x}{dt^2} + \omega_0^2x = 0$

where $\omega_0 = \sqrt{k/m}$ is natural angular frequency.

General Solution

$x(t) = A\cos(\omega_0 t + \phi)$

or equivalently:

$x(t) = C_1\cos(\omega_0 t) + C_2\sin(\omega_0 t)$

where $A$ is amplitude and $\phi$ is phase constant.

Velocity and Acceleration

$v(t) = \frac{dx}{dt} = -A\omega_0\sin(\omega_0 t + \phi)$

$a(t) = \frac{dv}{dt} = -A\omega_0^2\cos(\omega_0 t + \phi) = -\omega_0^2 x$

Maximum velocity: $v_{max} = A\omega_0$

Maximum acceleration: $a_{max} = A\omega_0^2$

Energy in SHM

Kinetic energy: $KE = \frac{1}{2}mv^2 = \frac{1}{2}mA^2\omega_0^2\sin^2(\omega_0 t + \phi)$

Potential energy: $PE = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\cos^2(\omega_0 t + \phi)$

Total energy: $E = KE + PE = \frac{1}{2}kA^2 = \text{constant}$

(Using $k = m\omega_0^2$)

Initial Conditions

Given $x_0 = x(0)$ and $v_0 = v(0)$:

$A = \sqrt{x_0^2 + \frac{v_0^2}{\omega_0^2}}$

$\tan\phi = -\frac{v_0}{\omega_0 x_0}$

Period and Frequency

Period: $T = \frac{2\pi}{\omega_0} = 2\pi\sqrt{\frac{m}{k}}$

Frequency: $f = \frac{1}{T} = \frac{\omega_0}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Simple Pendulum

For small angles ($\sin\theta \approx \theta$):

$\frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0$

$\omega_0 = \sqrt{\frac{g}{L}}$

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

Physical Pendulum

Extended object rotating about pivot:

Torque: $\tau = -mgd\sin\theta \approx -mgd\theta$

where $d$ is distance from pivot to center of mass.

$I\frac{d^2\theta}{dt^2} = -mgd\theta$

$\omega_0 = \sqrt{\frac{mgd}{I}}$

$T = 2\pi\sqrt{\frac{I}{mgd}}$

Torsional Pendulum

Restoring torque: $\tau = -\kappa\theta$

where $\kappa$ is torsional constant.

$I\frac{d^2\theta}{dt^2} = -\kappa\theta$

$\omega_0 = \sqrt{\frac{\kappa}{I}}$

Two-Body Oscillator

Two masses $m_1$ and $m_2$ connected by spring with constant $k$:

Reduced mass: $\mu = \frac{m_1m_2}{m_1 + m_2}$

$\omega_0 = \sqrt{\frac{k}{\mu}}$

System oscillates as if single mass $\mu$ attached to spring $k$.

Vertical Spring

Mass hanging from spring:

Equilibrium: $kx_{eq} = mg$

Displacement from equilibrium: $y = x - x_{eq}$

$m\frac{d^2y}{dt^2} = -ky$

Same SHM equation! Period independent of gravity:

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