Simple Harmonic Motion

Conditions for SHM

SHM occurs when the restoring force is proportional to displacement: $F = -kx$ This is Hooke's Law for a spring, where $k$ is the spring constant.

Key Quantities

• Amplitude ($A$): maximum displacement from equilibrium.
• Period ($T$): time for one complete cycle.
• Frequency ($f$): cycles per second; $f = 1/T$.
• Angular frequency: $\omega = 2\pi f = 2\pi/T$.

Position, Velocity, Acceleration

$x(t) = A\cos(\omega t + \phi)$ $v(t) = -A\omega\sin(\omega t + \phi)$ $a(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2 x$

• At $x = \pm A$: velocity is zero, acceleration is maximum.
• At $x = 0$: velocity is maximum ($v_{\max} = A\omega$), acceleration is zero.

Period of a Mass-Spring System

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

• Period depends on mass and spring constant, not amplitude.

Period of a Simple Pendulum

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

• Period depends on length and gravitational acceleration, not mass or amplitude (for small angles).

Energy in SHM

$E_{\text{total}} = \frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Energy oscillates between kinetic and potential, but total mechanical energy is constant (no friction).

Problem: A $0.5\ \text{kg}$ mass on a spring ($k = 200\ \text{N/m}$) is displaced $0.1\ \text{m}$ from equilibrium and released. Find the period, maximum speed, and maximum acceleration.

Solution:

Period: $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/200} = 2\pi\sqrt{0.0025} = 2\pi(0.05) \approx 0.314\ \text{s}$

$\omega = 2\pi/T = 20\ \text{rad/s}$

Max speed: $v_{\max} = A\omega = 0.1(20) = 2\ \text{m/s}$

Max acceleration: $a_{\max} = A\omega^2 = 0.1(400) = 40\ \text{m/s}^2$

• SHM occurs when the restoring force is proportional to displacement ($F = -kx$).
• The period of a spring depends on $m$ and $k$; for a pendulum it depends on $L$ and $g$.
• Energy oscillates between kinetic and potential; total energy $\propto A^2$.
• At equilibrium: max speed, zero acceleration. At maximum displacement: zero speed, max acceleration.