Vardaan Institute - Detailed Solutions: Master Sheet (Oscillations)

Section A: Kinematics Solutions

$x(t) = 5 \cos\left(2\pi t + \frac{\pi}{4}\right)$. Find $A, f, T, \phi_0$.

Sol: Amplitude $A = 5\text{ m}$. Angular frequency $\omega = 2\pi\text{ rad/s} \implies f = 1\text{ Hz}, T = 1\text{ s}$. Initial phase $\phi_0 = \pi/4\text{ rad} = 45^\circ$.

Derive kinetic and potential energy expressions in SHM.

Sol: $v = \omega \sqrt{A^2 - x^2} \implies K = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 (A^2 - x^2)$. Restoring force $F = -m\omega^2 x \implies U = \int_0^x m\omega^2 x dx = \frac{1}{2} m \omega^2 x^2$. $E = K + U = \frac{1}{2} m \omega^2 A^2$.

Section B: Pendulums & Springs Solutions

Derive time period of a simple pendulum $T = 2\pi\sqrt{L/g}$.

Sol: Restoring torque $\tau = -m g L \sin\theta \approx -m g L \theta$. Equation $I \alpha = -m g L \theta \implies (m L^2) \alpha = -m g L \theta \implies \alpha = -\frac{g}{L} \theta \implies T = 2\pi\sqrt{\frac{L}{g}}$.

Spring cut into two equal halves. Find new time period $T'$.

Sol: Cutting spring in half doubles spring constant $k' = 2k$. New time period $T' = 2\pi\sqrt{\frac{m}{k'}} = 2\pi\sqrt{\frac{m}{2k}} = \frac{T}{\sqrt{2}}$.

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