# SHM with two vertical springs

Consider a situation where there's two springs, basically like this

Except vertically. Turn the picture 90 degrees (I couldn't find a good picture quickly). So the top spring has spring constant $$k_1$$, bottom one $$k_2$$, both have natural length $$l$$, the total length in the picture is $$2l$$. Let's say we pull it down by $$x_0$$ and then release it. I hope I'm being clear!

So I chased the equations through and was surprised to find that the period of SHM seems to be $$2 \pi \sqrt \frac{m}{k_1 + k_2}$$

ie, independent of the initial conditions. And furthermore, the equilibrium point being at $$\frac{mg}{k_1+k_2}$$ below the middle point, again being independent again form the initial conditions - basically just depends how strong gravity is.

I found this pretty weird. Is it correct?

• Looks fine to me. I expect you've analysed a mass hanging from a single spring and noted that gravity displaces the equilibrium downwards but doesn't affect the period. There's not much difference with two springs except that the effective spring constant is greater because the pull of one spring and the push of the other co-operate! Oct 24, 2020 at 22:10
• A feature of the ideal spring at its equilibrium length is that it doesn't matter which side you attach it. (Put another way, tension and compression are exactly symmetric in elastic materials.) Therefore, this problem is equivalent to configuring two springs in parallel, which corresponds to adding the spring constants. Oct 26, 2020 at 19:55

For a single spring, the potential energy is:

$$V(z) = \frac 1 2 kz^2 + mgz$$

The equilibrium position is the root of:

$$\frac{dV}{dz} = kz + mg = 0$$

or:

$$z = -\frac{mg} k$$

which is below the gravity-free position of $$z=0$$.

If we transform to a new coordinate:

$$z' \equiv z + \frac{mg} k$$

then:

$$V(z') = \frac 1 2 k (z'- \frac{mg} k )^2 + mg(z'- \frac{mg} k )$$

$$V(z') = \frac 1 2 k(z'^2 - 2\frac{z'mg}k + (\frac{mg} k )^2) + mgz' -\frac{(mg)^2} k$$

$$V(z') =\frac 1 2 kz'^2 - \frac 1 2 \frac{(mg)^2} k$$

The quadratic term has the same pre-factor, so the frequency of oscillations is unchanged.

Note that the potential picked up a constant term. We can try to get rid of that by shifting our definition of zero potential energy from $$z=0$$ to $$z'=0$$ to account for the gravitational potential energy:

$$V' \rightarrow V + \frac{(mg)^2}{k}$$

so that:

$$V'(z') = \frac 1 2 kz'^2 + \frac 1 2 \frac{(mg)^2} k$$

What?! The term is still there, albeit with a different sign. This is the internal energy in the compressed spring.

Operating the spring about a new equilibrium point because of the addition of external field does not change the force about the equilibrium position, but it does change the internal energy of the system at the equilibrium position.

Regarding the fact that the effective spring constant in the two spring system is:

$$k_{\rm eff} = k_1 + k_2$$

You are adding the spring in parallel, which, like capacitors, add linearly.