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How can you determine the period of oscillation from a mass that is suspended from the ceiling? The equation becomes: ${{d^2x}\over dt^2}+kx-mg=0$. I am confused by the constant $mg$, because in the classical damped harmonic motion, the damping is velocity dependent.

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  • $\begingroup$ $kx - mg = k(x - mg/k)$ $\endgroup$ – Phoenix87 Jan 14 '15 at 16:05
  • $\begingroup$ There is no damping in a mass that is suspended from the ceiling, unless you take in account air resistance. $\endgroup$ – Mark Fantini Jan 14 '15 at 16:54
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This actually appears to be "ye olde harmonic oscillator" that has been shifted by a constant amount and not a damped harmonic oscillator that you state it is.

To see the shifted behavior, let $$ \eta(t)=x(t)-\frac{mg}{k} $$ Direct substitution gives you $$ \frac{d^2\eta}{dt^2}+k\eta=0\tag{1} $$ where we used that $mg/k$ is constant for the time-derivative term. (1) is your standard harmonic oscillator.

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  • $\begingroup$ This is very helpful. $\endgroup$ – Michael Angelo Jan 14 '15 at 18:51

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