# Finding phase angle of simple harmonic motion [closed]

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?

$x_{max}$ 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 $\varphi$.

• but what is x_max Feb 19 '12 at 1:11
• @Fendi, try using the advice Daniel gave you. If you can't figure out how to work out the problem given that, then you can come back and ask for clarification. Feb 19 '12 at 2:21
• Thats exactly why I posted back asking for clarification. I didn't get it. Feb 19 '12 at 3:04
• I've specified what $x_{max}$ is at the beginning of the post; it is the amplitude of the oscillations, i.e. the maximum displacement of the particle from its 0 position. If you mean how do you find the value of it; you use the initial conditions specified to find both $x_{max}$ and $\varphi$, the same way you'd find the values of any two equations with two unknowns. Feb 19 '12 at 7:50
• OFCOURSE !!! Simultaneous equations !!! Can you please clarify the following : Isn't the equation supposed to be x(0)=x_max cos (-phi) ? where phi is negative (wt - phi) ? and how do you get a w when you derive x(0)=x_max cos (-phi) ? is the derivative of phi the angular frequency ? Feb 19 '12 at 15:01