# Phase Angle for Simple Harmonic Motion

Phase Angle / Phase Question - Simple Harmonic Motion

## Homework Statement

If I were given a function of displacement for simple harmonic motion in the form of:

x = Acos($$\omega$$t + $$\phi$$)

Would the phase angle, $$\phi$$, always be the same? Say if I derived the equations into forms for velocity and acceleration as well. The phase angle would not change.. is that correct?

Also, I am confused about what is the phase and what is the phase angle? My textbook lists the "phase angle" as $$\phi$$ and some other sources list "phase" as ($$\omega$$t + $$\phi$$). What's the difference between these??

## Homework Equations

x = Acos($$\omega$$t + $$\phi$$)
v = -$$\omega$$Acos($$\omega$$t + $$\phi$$)
a = -$$\omega$$2Acos($$\omega$$t + $$\phi$$)

## The Attempt at a Solution

My belief is that it wouldn't change, but I want to be 100% sure.

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Hi, sorry to "bump" a thread but it's been a while now and I still want some clarity on this. Can anyone offer help?

As far as your first question is concerned, phase does not change if you take derivative of 'x' as long as the phase is independent of time.

ehild
Homework Helper
It is better to call Φ "the phase constant" It is not really angle, as no angle is involved in the SHM. Φ does not change if you calculate the time derivatives of the displacement. The phase is ωt+Φ. It depends on time. When x=Acos(ωt+Φ) has its maximum value the phase is 2kΠ (k is integer). And x=0 if ωt+Φ=(k+1/2)Π .

ehild