Oscillations (Calculus-Based)

The SHM Differential Equation

From $F = -kx = ma$: $\frac{d^2x}{dt^2} = -\omega^2 x$ where $\omega = \sqrt{k/m}$.

General solution: $x(t) = A\cos(\omega t + \phi)$

Deriving Period from the Differential Equation

• Mass-spring: $\omega = \sqrt{k/m}$, $T = 2\pi/\omega = 2\pi\sqrt{m/k}$
• Simple pendulum (small angle): $\omega = \sqrt{g/L}$, $T = 2\pi\sqrt{L/g}$
• Physical (compound) pendulum: $\omega = \sqrt{MgD/I}$, where $D$ is the distance from pivot to CM and $I$ is the moment of inertia about the pivot.

Torsional Oscillator

$\tau = -\kappa\theta \quad \Rightarrow \quad \omega = \sqrt{\kappa/I}$ where $\kappa$ is the torsion constant.

Energy in SHM

$E = \frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \text{constant}$

Using calculus: $v = dx/dt = -A\omega\sin(\omega t + \phi)$.

Damped Oscillations (Qualitative)

With a damping force $F = -bv$: $x(t) = A_0 e^{-bt/(2m)}\cos(\omega' t + \phi)$ where $\omega' = \sqrt{\omega_0^2 - (b/2m)^2}$ (slightly lower frequency).

• Underdamped: oscillates with decreasing amplitude.
• Critically damped: returns to equilibrium fastest without oscillating.
• Overdamped: slow return without oscillating.

Problem: A physical pendulum consists of a uniform rod of length $L$ and mass $M$ pivoted at one end. Find the period of small oscillations.

Solution:

Moment of inertia about the end: $I = \frac{1}{3}ML^2$.

Distance from pivot to center of mass: $D = L/2$.

$\omega = \sqrt{\frac{MgD}{I}} = \sqrt{\frac{Mg(L/2)}{\frac{1}{3}ML^2}} = \sqrt{\frac{3g}{2L}}$

$T = 2\pi\sqrt{\frac{2L}{3g}}$

• SHM arises from $d^2x/dt^2 = -\omega^2 x$; solution: $x = A\cos(\omega t + \phi)$.
• Physical pendulum: $\omega = \sqrt{MgD/I}$.
• Energy conservation provides an alternative to solving the differential equation.
• Damping reduces amplitude over time; frequency is slightly decreased.