Phase Angle for Simple Harmonic Motion

[mneox]

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(\omegat + \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 (\omegat + \phi). What's the difference between these??

Homework Equations

x = Acos(\omegat + \phi)
v = -\omegaAcos(\omegat + \phi)
a = -\omega²Acos(\omegat + \phi)

The Attempt at a Solution

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

Thanks for your clarification!

 

[mneox]

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?

 

[n.karthick]

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]

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