Consider a damped, driven harmonic oscillator:

$$m\frac{d^2x}{dt^2} + \beta\frac{dx}{dt} + kx = F(t) $$

I want to write this equation for time-varying $m$, $\beta$ and $k$ which are the mass, viscous damping coefficient and spring coefficient respectively.

In some papers, for the time-varying version of this equation the authors simply write:

$$m(t)\frac{d^2x}{dt^2} + \beta(t)\frac{dx}{dt} + k(t)x = F(t) $$

however this seems not quite right to me because according to Newton's second law, force equals to the time-derivative of momentum, then for time-varying mass:

$$F= \frac{d}{dt}\left(m\frac{dx}{dt}\right) = m\frac{d^2x}{dt^2} + \frac{dm}{dt}\frac{dx}{dt}$$

So there should be an extra term in the equation:

$$m(t)\frac{d^2x}{dt^2} + \left(\beta(t)+\frac{dm}{dt}\right)\frac{dx}{dt} + k(t)x = F(t) $$

I am not sure whether there should be extra terms for time-varying $\beta$ and $k$ too.

In short, I would like to know what time-dependent harmonic oscillator equation looks like.


1 Answer 1


I think your understanding of the matter is very much correct and the equation at the bottom is the one. Sooner or later - and maybe this is what the papers (silently?) do? - you will, for convenience, redefine $\beta$ to $\beta' = \beta + \frac{\text{d}m}{\text{d}t}$.

The way the model is built there will be no derivatives of products involving $\beta$ or $k$ and thus no extra terms for them.

  • $\begingroup$ So, there is no extra term coming from Hooke's law? $\endgroup$
    – trxrg
    Apr 5, 2022 at 16:33
  • $\begingroup$ I would appreciate more if you (sketchily) prove this result or give a reference. $\endgroup$
    – trxrg
    Apr 5, 2022 at 16:38
  • $\begingroup$ No, I mean, I think you understand the model, so this shouldn't be necessary. ;) But anyway: Friction force is a coefficient times velocity, to make it time-dependent make the coefficient time-dependent. What else? The force from Hooke's law is the product of a stiffness constant times position, to make it time-dependent... $\endgroup$
    – kricheli
    Apr 5, 2022 at 16:50
  • $\begingroup$ For example, if you change some of the conditions, say we cool down the spring to change its material properties, then the stiffness of the spring, i.e. $k$, changes as a function of time. How this change goes - the physics behind, detailed modelling - is a different matter, in the context of the equation above for the damped oscillator we can take it as being given as a function of time. $\endgroup$
    – kricheli
    Apr 5, 2022 at 16:54
  • $\begingroup$ On your example of spring cool-down, wouldn't it be easier if we model the change of k in detail and put it in the correct version of equation? Rather than re-modeling the behavior of the spring as a whole. $\endgroup$
    – trxrg
    Apr 5, 2022 at 17:37

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