Worked exercise — Finding A and φ from initial conditions

Problem

A mass $m = 0{,}2$ kg attached to a spring $K = 50$ N/m starts from $x_0 = 5$ cm from the equilibrium position with initial velocity $v_0 = -1$ m/s (towards equilibrium). Determine $A$, $\varphi$, $T$ and the position at $t = 0{,}1$ s.

Solution

Angular frequency and period. $\omega = \sqrt{K/m} = \sqrt{50/0{,}2} = \sqrt{250} \approx 15{,}8$ rad/s, hence $T = \frac{2\pi}{\omega} \approx 0{,}40 \ \text{s}$

Amplitude. Conservation of energy relates amplitude, position and initial velocity: $\tfrac{1}{2}K A^2 = \tfrac{1}{2}K x_0^2 + \tfrac{1}{2}m v_0^2$, hence $A^2 = x_0^2 + \frac{v_0^2}{\omega^2} = (0{,}05)^2 + \frac{(1)^2}{250} = 0{,}0025 + 0{,}004 = 0{,}0065$ $\ev{A \approx 0{,}081 \ \text{m} = 8{,}1 \ \text{cm}}$

Initial phase. From $x_0 = A\cos\varphi$ we get $\cos\varphi = 0{,}05/0{,}081 \approx 0{,}617$, hence $\varphi = \pm 51{,}9^\circ \approx \pm 0{,}906$ rad. The sign is decided by the velocity: $v_0 = -A\omega\sin\varphi$, and since $v_0 < 0$ we need $\sin\varphi > 0$, hence $\ev{\varphi \approx +0{,}906 \ \text{rad}}$

Position at $t = 0{,}1$ s. $x(0{,}1) = 0{,}081\cos(15{,}8\cdot 0{,}1 + 0{,}906) = 0{,}081\cos(2{,}49) \approx \ev{-0{,}065 \ \text{m}}$

Links

Topics: Oscillations and harmonic motion Concepts: Simple harmonic motion · Conservation of mechanical energy Skills: Symbolic set-up Objects: Spring