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I'm doing a problem on the simple harmonic motion but I didn't understand it really well.

The problem says: "A point describes a simple harmonic motion centered at the origin, the period is T=0.628s. For t = 0 point's position is x=0.15m etc..."

What it means with "harmonic motion is centered at the origin"? I thought that x=0 when t=0 but this is not right apparently.

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  • $\begingroup$ -1. Of course it is not right! The question states that $t=0$ when $x=0.15$. It cannot also be at $x=0$ when $t=0$. It cannot be at two places at the same time. $\endgroup$ – sammy gerbil Nov 14 '17 at 12:21
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The solution is symmetric about $x=0$. Thus, starting from $$ x(t)= x_0\cos(\omega t+ \phi) $$ you can derive $\omega$ and the phase $\phi$ from the data of your problem.

A solution not symmetric about the origin would be of the form $$ x(t)= A+ x_0\cos(\omega t+ \phi)\, . $$ The harmonic motion would then be "centered" about $A$.

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  • $\begingroup$ Oh I understand now! It was strange because the book didn't tackle no asymmetric SHM yet. Thank you a lot :) $\endgroup$ – Peto Nov 13 '17 at 16:33
  • $\begingroup$ @Peto this kind of language occurs because one could choose to have $x=0$ at the point where the sho is attached to a wall (if a spring-mass system). Then the equilibrium position would be at some position $\ne 0$. $\endgroup$ – ZeroTheHero Nov 13 '17 at 16:45
  • $\begingroup$ Yes it seems pretty legit. What I had understood is that the origin was the center of the axes ...but they're not the same in this kind of problems! $\endgroup$ – Peto Nov 13 '17 at 16:53
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It means that the SHM, in other words the body's oscillations such that its $a \propto x$, are oscillations that are about the origin. It does not necessarily imply that at $t = 0$, $x = 0$. For example, think of a pendulum which starts with from either the left or the right side to the bob's mean position. At $ t = 0$, it is starting from a position, $x \neq 0$.

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