Worked exercise — Vertical spring with energy balance

Problem

A mass $m = 0{,}5$ kg hanging from a vertical spring ($K = 200$ N/m) is pulled $3$ cm below the equilibrium position (already lowered by the weight) and released from rest. Determine: $T$, $A$, $v_\text{max}$, $a_\text{max}$, total energy.

Solution

Angular frequency and period. $\omega = \sqrt{\frac{K}{m}} = \sqrt{\frac{200}{0{,}5}} = 20 \ \text{rad/s}, \qquad T = \frac{2\pi}{20} \approx 0{,}314 \ \text{s}$

Why the weight doesn’t matter. The static equilibrium position already accounts for the weight ($K x_\text{eq} = mg$). Relative to that position, the oscillation is that of an ideal spring: the weight only shifts the centre of oscillation, it does not change its dynamics. So the amplitude is the imposed displacement, $A = 0{,}03 \ \text{m}$

Maximum quantities and energy. $v_\text{max} = A\omega = 0{,}03\cdot 20 = 0{,}6 \ \text{m/s}$ $a_\text{max} = A\omega^2 = 0{,}03\cdot 400 = 12 \ \text{m/s}^2$ $\ev{E = \tfrac{1}{2}K A^2 = 0{,}5\cdot 200\cdot (0{,}03)^2 = 0{,}09 \ \text{J}}$

Check. The same value is found as $\tfrac{1}{2}m v_\text{max}^2 = 0{,}5\cdot 0{,}5\cdot (0{,}6)^2 = 0{,}09$ J: the energy oscillates between elastic and kinetic form without losses.

Collegamenti

Argomenti: Oscillations and simple harmonic motion Concetti: Simple harmonic motion · Elastic potential energy · Conservation of mechanical energy Competenze: Conservation of energy Oggetti: Spring