# What will be different if the spring is not massless?

In almost all text, it uses the massless spring as example to illustrate the idea of elastic potential energy. I wonder what's really being changed if we consider the mass of the spring? I saw a problem on a text regarding the non-massless spring. It is said that the stretched spring (don't know the length) has uniform density and mass $M$, the elastic constant k is also known. If we hold it vertical so it stretches naturally due to the gravity, what's the elastic potential will be. I am trying to understand this from physical point of view as follows but I don't know if it is correct or not since there might be some knowledge beyond undergraduate course I need to solve the problem

1) I think the spring is actually has the gravity acting on it's center of mass, so if we choose the (gravitational) potential reference to be zero (at where the hanging point), the gravitational potential is $Mgh$ where $h$ is the where the center of mass is .

2) From 1), can we say that the elastic potential energy is actually the same as the gravitational potential energy, so $$\frac{k}{2}(\Delta l)^2 = Mgh$$

So if we figure out $h$, we know the elastic potential energy. But the problem how to find the center of mass. I bring this to some senior students, some of them said the center of mass will not be changed, so the center of mass when the spring is not stretched and when it is naturally stretched will be the same, that's is to say, when it is not stretched, the center of mass is $l_0/2$ so $h=l_0/2$. Well, frankly, I don't see the point.

Someone show me the math to calculate the center of mass, which is an integral $$h = l_c = \frac{1}{M}\int_0^L \frac{M}{l_0}xdx$$ there he said $L$ is the fully-extended length and $l_0$ is the unstretched length, I don't know how to do the math but I plug that into mathematica and it gives

$$h = l_c = \frac{L^2}{2l_0}$$

but if this is correct, we need to know the fully-extended length and original length to figure the center of mass. I don't think we get that information. So do we really need other knowledge to find the elastic potential energy for non-massless spring?

-
Related: physics.stackexchange.com/q/64934/2451 and links therein. – Qmechanic Sep 26 '13 at 23:41
That question is asking dynamics but in my case, I only want to see the static case, the spring is not even moving. – user1285419 Sep 27 '13 at 2:08
A moving non-massless spring will hold kinetic energy. If there is gravity then it will have potential energy also. – ja72 May 14 '14 at 17:35

In general you can think massive springs as being a sequence of massless springs and masses connected in series. So you can compute the energy in each of these pairs spring-mass and then sum up (integrate).

I would take,

$$E_p=\int_0^L\frac12 \frac{k}{L} \left(\Delta x (x)\right)^2 dx$$

where $\Delta x$ is the displacement for each $x$. You can calculate it statically using the force for a small piece of spring located at $x$:

$$F(x)=k.\Delta x(x)$$

and $F(x)=m(x) g$, where $m(x)$ is the mass of the spring until $x$ (imagine the spring hanging and $x$ begins in the top and go downwards). It holds that $m(x)=\mu x$, where $\mu$ is the linear density $M/L$. Then

$$E_p=\int_0^L\frac12 \frac{k}{L} \left(F (x)\right)^2 dx$$

$$E_p=\int_0^L\frac12 \frac{k}{L} \left(\frac{Mg}{kL}x\right)^2 dx$$

$$E_p=\frac12 \frac{kM^2g^2}{L^3k^2} \int_0^L x^2 dx$$

$$E_p=\frac12 \frac{kM^2g^2}{L^3k^2} \frac13 L^3$$

$$E_p=\frac{M^2g^2}{6k}$$

-

Hint: The change in position of 'centre of mass' is half of its extension. Give it a try then come back, I will update the answer.

-
Thanks for the hint, I will work it again. – user1285419 Sep 27 '13 at 2:26