Simple harmonic motion interpretation problem

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Discussion Overview

The discussion revolves around the interpretation of the expression for time in simple harmonic motion (SHM), specifically the relationship between displacement, amplitude, and time. Participants explore the implications of using cosine versus sine in the SHM equations and the conditions under which displacement is zero.

Discussion Character

  • Conceptual clarification, Debate/contested, Technical explanation

Main Points Raised

  • One participant expresses confusion about the expression ##t= \frac{1}{\omega} \cos^{-1}(x/A)##, particularly when ##x = 0##, questioning why it does not yield ##t = 0##.
  • Another participant notes that ##\cos^{-1}## has multiple roots, suggesting that this could lead to different interpretations of the time associated with a given displacement.
  • A further comment reiterates that ##\cos^{-1}## has multiple roots, emphasizing that one of these roots corresponds to the amplitude rather than zero displacement.
  • One participant suggests that the original equation assumes ##t=0## corresponds to maximum displacement, indicating that a sine function should be used if zero displacement is desired at ##t=0##.
  • Another participant agrees with this interpretation, clarifying that the displacement referred to in the equation is from the equilibrium position, not from the initial time.
  • A final participant acknowledges their misunderstanding and thanks others for the clarification.

Areas of Agreement / Disagreement

Participants generally agree on the need to clarify the use of cosine versus sine in the context of SHM equations, but there remains some uncertainty regarding the implications of multiple roots of the cosine function and the initial conditions of the motion.

Contextual Notes

The discussion highlights potential limitations in understanding the initial conditions of SHM and the dependence on the choice of trigonometric function in the equations used.

Celso
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I'm in trouble trying to understand the expression ##t= \frac{1}{\omega} cos^{-1}(x/A)## that comes from ##x = Acos(\omega t)##, in which ##A## is the amplitude, ##t## is time and ##x## is displacement.
When ##x = 0##, ##t = \frac{\pi}{2\omega} ##, shouldn't it be 0 since there was no movement?
 
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##\cos^{-1}## has multiple roots.
 
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Dale said:
##\cos^{-1}## has multiple roots.
It has a root in ##x/A = 1##, but in that case the distance would be equal to the amplitude, not zero
 
You seem to be using a form of the SHM equation that treats ##t=0## as a time when ##x## is a maximum. If you want ##x=0## at ##t=0## you need to use ##\sin##, not ##\cos##.
 
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Yes, @Ibix is right. In this equation x is the displacement from equilibrium, not the displacement from t=0. It starts at the peak.
 
ah that's my mistake, thank you guys
 

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