# Massive spring

dEdt
My textbook derived an expression for the frequency of oscillation of a mass m attached horizontally to a spring with mass M and constant k. Assuming that M<<m, it got that
$$\omega^2 = \frac{k}{m + M/3}$$.
The author used the conservation of energy to get this expression, but he assumed that the total potential energy still still satisfies $$U = \frac{1}{2} k x^2$$, where x is the extension of the spring. Why is this formula still correct?

Also, I wanted to derive the same result using forces, but I'm having some trouble. How would I go about this? Thanks a lot!

Homework Helper
Because M << m, you can assume the inertia forces on the spring's own mass M do not affect the deformation in the spring. In other words, if you measure from the fixed end, the spring is length L, and the motion of the large mass m is A sin w t, then the motion of any point the length of the spring is
u(x,t) = A (x/L) sin wt

This is the same as for a massless spring, so the potential energy is the same.

The M/3 term comes from the extra kinetic energy of the mass M. You can find that by integrating along the length of the spring.

This is only an approximate solution, because the inertia of the spring does affect the displacments along its length, and the equation for u(x,t) given above ignores that fact.

Homework Helper
hi dEdt!

(have an omega: ω )
… he assumed that the total potential energy still still satisfies $$U = \frac{1}{2} k x^2$$, where x is the extension of the spring. Why is this formula still correct?

it's always correct (for the PE)

the only relevance of M << m is that he's ingoring the KE of the spring
Also, I wanted to derive the same result using forces, but I'm having some trouble. How would I go about this? Thanks a lot!

you should get the integral of the energy equation

show us what you did get

dEdt
Here's what I (tried) to do with the force approach:
The external force on the spring-mass system is the force from the wall on the spring F_w it is attached to. So $$F_w = (M + m) \frac{d^2}{dt^2}x_{cm}$$, where x_cm is the position of the center of mass. Subbing in $$x_{cm} = \frac{mx + Mx/2}{m+M}$$ (x is now measured from wall) and $$F_w = k \Delta x$$ gives that
$$\omega^2 = \frac{k}{m+M/2}$$. :S

Homework Helper
hi dEdt!
… I wanted to derive the same result using forces …

the trouble with using forces is that the two forces at opposite ends of the spring are not the same …

you're used to the tension or compression in a rod (or a spring) being the same at both ends, but that's only if the centre of mass isn't moving

in this case, you cannot say Fw = k∆x, you'd have to find equations for Fw and Fm, use Fw - Fm = Mx''/2, and eliminate the Fs

now do you see why it's so much easier to use energy?

dEdt
Well, I thought that because the mass of the spring was small, I could approximate that the forces on the two ends of the spring were the same.

At any rate, if I go about it your way, I still get the wrong answer! I get that $$\omega^2 = \frac{k}{m+M}$$.

The only possible resolution to this contradiction would be that the force of the spring on the mass is not $$-k\Delta x$$. This would make sense, because if not, the period of the mass shouldn't be different than when it's massless because the force doesn't change. But then what's Hooke's Law for a massive spring????

Homework Helper
hi dEdt!

hmm … you've obviously used 1/2 mv2 + 1/2 Mv2 + 1/2 kx2 = constant

that's wrong because the kinetic energy of the spring is not 1/2 Mv2

that would be correct only if the whole spring had velocity v …

you'll have to integrate (over the length of the spring) to find what the correct KE is

dEdt
Sorry Tiny-tim, I botched my last post.

What I meant to say when I said that I would try your way was that I would use Fw - Fm = Mx''/2, which gives me the omega I wrote before. Sorry for the confusion (with energy, I did, following my textbook, use 1/6Mv2).

So, does Hooke's Law change with a massive spring?

Homework Helper
no, Hooke's Law is still the same …

it only concerns the force, and has nothing to do with the mass or inertia

(and for example the spring energy is totally separate from the kinetic energy, and indeed from gravitational potential energy)

dEdt
Alright, by why doesn't the force approach work then? We have
$$F_m - F_w = \frac{Mx''}{2}$$, $$-F_w = (m+M)(x_{CM})''$$, and $$F_m = kx$$.

Combining these together, and using that $$x_{CM}'' = (mx +Mx/2)/(m+M)$$, I get a wrong expression for the angular frequency. I'm confused, help! :(

Homework Helper
hmm … i think the problem is that tension in a massive spring varies along its length (quadratically, i believe) …

but i don't know the formula for it …

perhaps you can work it out?

dEdt
Should that matter? wouldn't the internal forces just cancel?

I think the problem is with Hooke's Law, personally

Alvasin
Spring usually use to absorb the energy.Where it is needed to absorb huge amount of energy there usually use massive spring.Their absorbing power is outstanding.energy absorb equation can be written as-

ω = [k/(M + m/3 + o(m2/M) )]1/2.