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.

  • $\begingroup$ $kx - mg = k(x - mg/k)$ $\endgroup$
    – Phoenix87
    Commented Jan 14, 2015 at 16:05
  • $\begingroup$ There is no damping in a mass that is suspended from the ceiling, unless you take in account air resistance. $\endgroup$ Commented Jan 14, 2015 at 16:54

1 Answer 1


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.

  • $\begingroup$ This is very helpful. $\endgroup$ Commented Jan 14, 2015 at 18:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.