Do springs and projectiles under gravity exist in the same family?

Question

Background

A 1D vertical spring subject to gravity satisfies Hooke's Law:

$$m x''(t) = -kx(t) + g$$

where $m$ is the mass at the end of the spring, $x(t)$ is the position of the mass at time $t$, $x''(t)$ is its acceleration, $k$ is the stiffness coefficient, and $g$ is the gravitational force.

Solving this ODE (I used Maple, shame on me), I get:

$$x \left( t \right) =\sin \left(\frac{\sqrt{k}t}{\sqrt {m}}\right)c_1+\cos \left(\frac{\sqrt {k}t}{\sqrt {m}}\right)c_2 + {\frac {g}{k}},$$

where $c_1$ and $c_2$ are integration constants.

If my mass were detached from the spring, then this amounts to setting $k=0$ and the mass falls according to:

$$mx''(t) = g$$

and solving this produces–as expected–a quadratically increasing displacement:

$$x(t) = \frac{g}{2m}t^2 + c_3 t + c_4$$

where $c_3$ and $c_4$ are integration constants.

With initial conditions: We can simplify this problem further by assuming homogenous initial conditions in both cases: $x(0) = x'(0) = 0$. For the spring solution we get:

$$x(t) = -\frac{g}{k}\cos \left(\frac{\sqrt{k}}{\sqrt{m}}t \right) + \frac{g}{k}$$

and for the falling object we get:

$$x(t) = \frac{g}{2m}t^2.$$

Question

Because of the divide by $k$ I cannot simply consider $k=0$ to get from the spring solution to the falling object solution. Naively taking the limit seems to produce $x(t) = \infty$.

Is there a correct way to take the limit or to solve the ODE so that I get a solution that transitions from the oscillatory spring into the falling object as $k\rightarrow 0$?

Note: the conversation in the comments here mention similar behavior but without a solution to the specific question.

Bonus points for convincing maple to do this since my actual problem is a more complicated of this simple scenario.

Answer 1

Let's consider the case where the initial conditions are $$ x(0) = 0, \ \ \ \ \ \ \dot x(0) = 0.$$ The solution to your coupled spring + gravity equation is $$ x(t) = \frac{g}{k} \left[ 1 - \cos \left( \sqrt{\frac k m} t\right) \right].$$ (I fixed the missing minus sign in the equation as pointed out by Ross.)

Now suppose we send the spring constant $k$ to zero. Then for "moderate" values of $t$ (more precisely, for $t \ll \sqrt{\frac m k}$ - remember that $\sqrt{\frac m k }$ is very large because $k$ is small), we can approximate the cos with the first few terms of its Taylor series: $$ x(t) \approx \frac{g}{k} \left[ 1 - 1 + \frac 1 {2!} \left( \sqrt{\frac k m} t \right)^2 \right] = \frac{gt^2}{2m}.$$ And this is precisely the solution to the gravity-only equation with initial conditions $x(0) = 0$ and $\dot x (0) = 0$.

Answer 2

You need a minus sign on your $kx(t)$ which will make your solutions sinusoidal instead of exponentially growing. You will then have terms in $\cos \sqrt {\frac km}t$ and $\sin \sqrt{\frac km}t$ The reason you get $\frac gk$ is because the right place to measure $x$ from is the equilibrium position where the spring force balances gravity. You are measuring $x$ from the natural point where the spring force is zero. As $k \to 0$ the balance position gets farther and farther from the natural point of the spring in order to make $mg=kx$. Of course you can put the zero anywhere you want, but you get that $\frac gk$ term when you put it where you do. As $k$ gets small, another reasonable thing is to imagine it is zero and use your $k=0$ solution and see the spring as a small perturbation. I would switch over to this when you stop having oscillatory behavior because there isn't room or time for you to complete a cycle.