Simple harmonic motion interpretation problem

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SUMMARY

The discussion centers on the interpretation of the simple harmonic motion (SHM) equation, specifically the expression t = \frac{1}{\omega} \cos^{-1}(x/A) derived from x = A \cos(\omega t). Participants clarify that when x = 0, the correct time should be t = \frac{\pi}{2\omega}, indicating a quarter period into the motion, not zero. The confusion arises from treating t=0 as a maximum displacement, whereas x=0 at t=0 requires using the sine function instead of cosine. This distinction is crucial for accurately modeling SHM.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM) principles
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Knowledge of angular frequency (\omega) in oscillatory motion
  • Ability to manipulate inverse trigonometric functions, such as \cos^{-1}
NEXT STEPS
  • Study the derivation and applications of the sine function in SHM
  • Explore the implications of phase shifts in harmonic motion
  • Learn about the graphical representation of SHM and its key characteristics
  • Investigate the role of angular frequency (\omega) in oscillatory systems
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone seeking to deepen their understanding of simple harmonic motion and its mathematical representations.

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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