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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2$\begingroup$ Let $x = y+\frac{k}{\omega^2}$... $\endgroup$– By SymmetryCommented Dec 4, 2017 at 14:24
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1 Answer
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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. $\endgroup$– ymufCommented Dec 4, 2017 at 15:02
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$\begingroup$ @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. $\endgroup$– GertCommented Dec 4, 2017 at 15:16
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$\begingroup$ Ah... sorry, I didn't understand the example before. Thank you. $\endgroup$– ymufCommented Dec 4, 2017 at 15:20