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This question already has an answer here:

Would

$$ \frac{{\rm d}^2 x}{{\rm d}t^2} + \omega^2 x = {\rm (constant)}$$ be an SHM?

My instinct is no, as acceleration would not be zero at x = 0, but a solution in my book uses this equation to find angular frequency for an SHM.

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marked as duplicate by sammy gerbil, Community Dec 5 '17 at 13:05

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    $\begingroup$ Let $x = y+\frac{k}{\omega^2}$... $\endgroup$ – By Symmetry Dec 4 '17 at 14:24
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Say: $$\ddot{x}+\omega^2 x=C$$ $$\ddot{x}+\omega^2 x-C=0$$

Substitute:

$$u=\omega^2 x-C$$ $$\dot{u}=\omega^2\dot{x}$$ $$\ddot{u}=\omega^2\ddot{x}\implies \ddot{x}=\frac{1}{\omega^2}\ddot{u}$$ $$\implies \frac{1}{\omega^2}\ddot{u}+u=0$$ Or:

$$\ddot{u}+\omega^2 u=0$$

Which is the ODE for an SHM.

Solve, apply boundary conditions and backsubstitute $u$.

Here's a practical example, fully worked.

My instinct is no, as acceleration would not be zero at $x = 0$

That's not a requirement of an SHM, see my worked example.

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  • $\begingroup$ I still don't understand why acceleration doesn't have to be zero at the mean position. Mean position implies stable equilibrium which implies that the net force at that point is zero. $\endgroup$ – ymuf Dec 4 '17 at 15:02
  • $\begingroup$ @user165756: did you check my worked example (link)? As $ma=F-kx$, at $x=0$, clearly $a\neq 0$. Mean position would be stable equilibrium if NO oscillation took place at all. $\endgroup$ – Gert Dec 4 '17 at 15:16
  • $\begingroup$ Ah... sorry, I didn't understand the example before. Thank you. $\endgroup$ – ymuf Dec 4 '17 at 15:20

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