# Massless spring

1. Jun 5, 2009

### peeyush_ali

can we exert 2 unequal forces in opposite directions on the two sides of an ideal<mass less> spring?

2. Jun 5, 2009

### Staff: Mentor

Not in a steady state, no.

3. Jun 5, 2009

### peeyush_ali

can u tell me some more facts abt springs..how to solve problems involving springs with mass and without mass..??

4. Nov 13, 2009

### YorkLarry

A massless spring is compressed then suddenly released. Does it overshoot its uncompressed length? Where does the stored energy go?

5. Nov 13, 2009

### niu ren

It can not overshoot its uncompressed length,because it's not accord with conversation of energy. when you release the massless spring,the stored energy switch the spring's kinetic energy,or else why the spring can kinetic without energy

6. Nov 16, 2009

### YorkLarry

I did some calculations after the last post. I've never seen an actual massless spring, and I don't know how to make one or where to buy one, so I added some mass m by gluing a mass m/2 to each end of the spring. I worked out the dynamics and then took the limit as m approaches zero.

If the spring with relaxed length Lo is compressed to length L1 and then released, it will oscillate with amplitude (Lo - L1) independent of m. It has angular frequency sqrt(k/m). The potential and kinetic energy exchange during the motion so as to keep the total energy constant. The maximum kinetic energy equals the maximum potential energy.

As the limit of small m is approached, the angular frequency increases, and the ends of the spring move faster and faster. So the spring will overshoot its uncompressed length regardless of how small m is. The stored energy goes into kinetic energy during the motion, except at the extremes of spring length.

If the mass m is distributed along the length of the spring instead of being concentrated at the ends, the argument is the same. Slices of mass other than at the ends move more slowly than the ends so the kinetic energy formulas are slightly different. The amplitude is still Lo - L1. The angular frequency is still proportional to sqrt(k/m) with the proportionality constant dependent on details of the mass distribution. Behavior in the massless limit is the same.

I can provide details if anyone is interested.

7. Dec 18, 2009

### YorkLarry

Apologies re the previous post - the angular frequency should be 2 sqrt(k/m). I dropped the factor of two.