Finding Phase angle of Simple Harmonic Motion?

Question

A sinusoidal oscillator has :

$$x=x_{max} \cos(\omega t - \varphi )$$

Period is 2, initial displacement is 100mm initial velocity is 200mm/s

What is the phase angle assuming $-\pi < \varphi < \pi$

How do I go about solving this?

Is the phase $(\omega t - \varphi)$? But I do not know what $x_{max}$ is, how am I supposed to solve for the angle?

Answer 1

$x_{max}$ is the amplitude of the oscillations, and yes, ${\omega}t - \varphi$ is the phase.

We know that the period $T$, is the reciprocal of the frequency $f$, or $$T = 1/f$$

We also know that $\omega$, the angular frequency, is equal to $2\pi$ times the frequency, or $$\omega = 2{\pi}f$$

From here, we can use the initial conditions to find the amplitude.

$x(0) = x_{max}cos(\varphi)$

$\dot{x}(0) = {\omega}x_{max}sin(\varphi)$

From here it should be a simple matter to find $\varphi$.