Behaviour of harmonic oscillator when elastic constant$\to +\infty$.

Question

I have a silly question about asymptotics and physical meaning. Suppose you have this equation here

\begin{align*} x'' + \dfrac{k}{m}\,x &= -g \\ x(0) &= x_0 \\ x'(0) &= v_0. \end{align*}

Assuming I did all my calculations right the solution is something like

$$ x(t) = \left(x_0 - \frac{m}{k}\,g\right) \cos\left(\sqrt{\frac{k}{m}}\,t \right) + \sqrt{\frac{m}{k}}\,v_0 \sin\left(\sqrt{\frac{k}{m}}\,t\right) - \frac{m}{k}\,g. $$

Now if I let $k\to +\infty$ I get $x(t) \sim x_0 \cos\left(\sqrt{\dfrac{k}{m}}\,t \right)$ which is kind of unexpected in some sense and I'd like to know if the interpretation I am giving is correct:

1. If $x_0 = 0$ then the system is not affected by the gravity at all, initial velocity won't matter and there's only one dominant term in the asymptotic expansion.
2. If $x_0 \neq 0$ the system will oscillate indefinitely, this bit was not really expected.

I was expecting to see the system completely still no matter what with an extremely stiff spring.

Is my interpretation correct?

Answer 1

To understand what occurs when you let $k\to\infty,$ I would divide your DE through by $k:$ $$\frac{x''}{k} + \frac{1}{m}\,x = -\frac{g}{k}.$$ Then you can see that the far left term and the RHS go to zero, and you're left with $\dfrac{x}{m}=0,$ or $x=0,$ which is what you'd expect with an infinitely stiff spring.