2

Suppose the equation of motion for some oscillating system takes the following form:

$$\ddot{\theta}+\dot{\theta}^2\sin\theta+k^2\theta\cos\theta=0$$

Applying small angle approximation to $\theta$ gives $\sin\theta\approx\theta$ and $\cos\theta\approx1$,

$$\ddot{\theta}+\dot{\theta}^2\theta+k^2\theta=0$$

But I am wondering if it is alright to set $\dot{\theta}^2=0$ so that the equation simplifies to

$$\ddot{\theta}+k^2\theta=0$$

If so, how can this approximation be justified?

Qmechanic
  • 201,751
  • 1
    Your approximation is only true if that physically is the case, however $\theta \sim 0$ as you know doesn't imply $\dot{\theta}\sim 0$ – Triatticus Feb 06 '23 at 07:16
  • 1
  • some oscillating system* Which oscillating system? What physical reality underpins the DE? Try solve the DE with wolfram alpha.
  • – Gert Feb 06 '23 at 10:20
  • 2
    You could rescale $\theta\rightarrow\epsilon\theta$ and expand your initial DE in orders of the small amplitude $\epsilon$, s.t. your final equation is the original DE up to order $\epsilon^2$, the next non-vanishing order beeing $\mathcal{O}(\epsilon^3)$. – go_science Feb 06 '23 at 14:08
  • The origin (in velocity-phase space) is an equilibrium point of your ODE, which is why it makes sense to consider expanding the system about the equilibrium solution. Of course if you only want to consider up to first order, then you can drop several terms; see here for a more detailed description of what it means to expand about equilibrium points. But keep in mind that this means you're only solving a linearized problem about the equilibrium solution. – peek-a-boo Feb 06 '23 at 18:58