Simple harmonic motion interpretation problem

[Celso]

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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[Dale]

$\cos^{-1}$ has multiple roots.

 

[Celso]

$\cos^{-1}$ 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

 

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[Ibix]

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 $\cos$.

 

[Dale]

Yes, @Ibix is right. In this equation x is the displacement from equilibrium, not the displacement from t=0. It starts at the peak.

 

[Celso]

ah that's my mistake, thank you guys