Phase Angle for Simple Harmonic Motion

  • Thread starter mneox
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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([tex]\omega[/tex]t + [tex]\phi[/tex])

Would the phase angle, [tex]\phi[/tex], 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 [tex]\phi[/tex] and some other sources list "phase" as ([tex]\omega[/tex]t + [tex]\phi[/tex]). What's the difference between these??

Homework Equations



x = Acos([tex]\omega[/tex]t + [tex]\phi[/tex])
v = -[tex]\omega[/tex]Acos([tex]\omega[/tex]t + [tex]\phi[/tex])
a = -[tex]\omega[/tex]2Acos([tex]\omega[/tex]t + [tex]\phi[/tex])

The Attempt at a Solution



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

Thanks for your clarification!
 
Last edited:

Answers and Replies

  • #2
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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?
 
  • #3
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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.
 
  • #4
ehild
Homework Helper
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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
 

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