Massive spring

  • Thread starter dEdt
  • Start date
  • #1
288
2

Main Question or Discussion Point

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
[tex]\omega^2 = \frac{k}{m + M/3}[/tex].
The author used the conservation of energy to get this expression, but he assumed that the total potential energy still still satisfies [tex]U = \frac{1}{2} k x^2[/tex], 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!
 

Answers and Replies

  • #2
AlephZero
Science Advisor
Homework Helper
6,994
291
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.
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,832
249
hi dEdt! :smile:

(have an omega: ω :wink:)
… he assumed that the total potential energy still still satisfies [tex]U = \frac{1}{2} k x^2[/tex], 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 :wink:
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 :confused:

show us what you did get :smile:
 
  • #4
288
2
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 [tex]F_w = (M + m) \frac{d^2}{dt^2}x_{cm}[/tex], where x_cm is the position of the center of mass. Subbing in [tex]x_{cm} = \frac{mx + Mx/2}{m+M}[/tex] (x is now measured from wall) and [tex]F_w = k \Delta x[/tex] gives that
[tex]\omega^2 = \frac{k}{m+M/2}[/tex]. :S
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
249
hi dEdt! :smile:
… 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 :wink:

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 :redface:

now do you see why it's so much easier to use energy? :biggrin:
 
  • #6
288
2
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 [tex]\omega^2 = \frac{k}{m+M}[/tex].

The only possible resolution to this contradiction would be that the force of the spring on the mass is not [tex]-k\Delta x[/tex]. 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????
 
  • #7
tiny-tim
Science Advisor
Homework Helper
25,832
249
hi dEdt! :smile:

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 :wink:
 
  • #8
288
2
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?
 
  • #9
tiny-tim
Science Advisor
Homework Helper
25,832
249
no, Hooke's Law is still the same …

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

(and for example the spring energy is totally separate from the kinetic energy, and indeed from gravitational potential energy)
 
  • #10
288
2
Alright, by why doesn't the force approach work then? We have
[tex] F_m - F_w = \frac{Mx''}{2}[/tex], [tex]-F_w = (m+M)(x_{CM})''[/tex], and [tex]F_m = kx[/tex].

Combining these together, and using that [tex]x_{CM}'' = (mx +Mx/2)/(m+M)[/tex], I get a wrong expression for the angular frequency. I'm confused, help! :(
 
  • #11
tiny-tim
Science Advisor
Homework Helper
25,832
249
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? :wink:
 
  • #12
288
2
Should that matter? wouldn't the internal forces just cancel?

I think the problem is with Hooke's Law, personally
 
  • #13
10
0
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.
 

Related Threads for: Massive spring

Replies
5
Views
2K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
4
Views
1K
  • Last Post
Replies
15
Views
746
Replies
4
Views
846
Replies
15
Views
3K
Replies
3
Views
2K
Top