How is energy converted when massless spring is attached to a rigid support

Say a massless spring of length $l$ is attached to a rigid support. It is extended to length $l + x$. Now at this position, the force extending it is removed. How will its potential energy be converted ? What happens ?

• The potential energy is actually elastic energy stored in the spring. The spring will go back to rest to its original length (after a few "contortions"), and the elastic energy will be lost as heat.
– user65081
Oct 26, 2014 at 8:52

The idea of a massless spring is sometimes useful as is a spring whose mass is much, much less than the other masses within the system under consideration.

In this case a massless spring implies that there will be an infinite acceleration when the stretching force is released so better to consider a spring with some mass.

First assume that there are no frictional forces acting.
The extended spring will have forces within it which will reduce its extension and during this process the elastic potential energy is converted to kinetic energy.
When the spring reaches its unextended length it will have no elastic potential energy but will have kinetic energy and so will overshoot the equilibrium position.
Now the forces acting on the compressed spring will be such as to try and extend it and the kinetic energy of the spring will be converted to elastic potential energy.
This continues until the spring stops moving but has a store of elastic potential energy.
The cycle is reversed and the spring undergoes oscillatory motion.

If there is a small amount of air resistance/friction (damping) then the amplitude of successive oscillations will decrease and eventually the spring will stop moving.
If there is a great deal of damping then the spring may stop moving when it reaches its unextended length without ever undergoing any oscillatory motion.

So in the real world all that elastic potential energy that the stretched spring had eventually becomes heat.

As Julian said, it is the Elastic Potential Energy that is stored in a spring, which is the potential energy of an elastic object that is deformed under tension (extension or pulling of the spring) or compression (compression or pushing of the spring). It comes into play because of a force ($~\text{EM}$ force between the constituent particles) that tries to restore the object to their original shape. If the stretch is released, most of the energy is transformed into $E_{k}$ while some energy will get converted into heat.

• To calculate the Elastic Potential Energy, we can apply the formula $\frac{1}{2}kX^2$ ; here $X$ is the displacement of the linear spring from its mean position, and $k$ is is a constant factor property of the spring.

• We know by Hooke's law that $$F\propto X~;~~~\therefore F=kX$$ Since force increases linearly with x , the average force that must be applied is $$F=\frac{F_0+F_X}{2}=\frac{1}{2}kX$$ $$\therefore ~~E_{p~\text{(spring)}}=F\dot{}s=\frac{kX^2}{2}$$

• What sort of velocities do the parts of the massless spring achieve as they temporarily store the energy of the system as kinetic energy? Oct 26, 2014 at 14:29
• Oh I forgot to take into account that. If the spring is massless, and we are not taking relativistic concepts into consideration, energy should practically be zero.
– user49111
Oct 26, 2014 at 15:55
• Typically one uses the approximation that a spring is massless to simplify calculations of a mass and spring problem. With this approximation, you can ignore the kinetic energy of the spring and just calculate that of the mass. The approximation is only good when the mass of the mass is much larger than the mass of the spring. So your question is what happens when the approximation is bad. The answer is your approximation gives the wrong answer. Jul 30, 2015 at 13:25

The are no means to hold kinetic energy in MASSLESS spring. So, the potential energy of extension, will instantly converted to infinite speed of contracting spring.

If spring is absolutely rigid itself, it will contract to equivalent minimal length and will start to extending back. Once it extends to initial length, it will start to contract back, and so on. It will oscillate forever.

The frequency of oscillations will be infinite too.

After the elongation of x units in the spring, the spring develops energy in the form of spring compression or elongation energy(the work done in elongation is converted into spring energy), spring tries to get into its mean position. Now, if the system is isolated the spring will perform SHM otherwise friction and other forces will make it stop its motion. The energy spring develops can be described as, E(spring) = 1/2*k*(x^2).