# What will happen if I just stretch a spring and release it?

What will happen if I stretch a spring and release it. There is no mass in the spring and there is no mass attatched to the spring. Will it stop when it aquires its natural length or will it continue to move? If it stops at the natural length, how will its energy will remain conserved? Any help will be appreciated. Thanks

• A massless spring is unphysical. That is to say, the laws of kinematics will give you meaningless answers if you try to analyze it. Depending on what you really want to know (since you will never use an actual massless spring in real life), you might find the concept of limits to be useful. I.e., instead of asking about a massless system, you ask, how does the behavior of a massive system change as the mass shrinks, approaching zero? – Solomon Slow Nov 30 '18 at 18:55

You can't really say it's got no mass without leading to some pretty absurd conclusions. Suppose the spring's free length is $$L_0$$, its spring constant is $$k$$ and uniform mass $$m$$ (which we shall tend to zero eventually). Suppose the spring is fixed to a wall in one of its ends, for simplicity, and assume that it compresses and extends uniformly to a length $$L$$.

The elastic potential energy can be just written down without much thought:

$$U=\frac{1}{2}k(L-L_0)^2$$

The kinetic energy is a bit more complicated. Let $$v(x)$$ be the velocity of the spring along its length as a function of the distance $$x$$ from the wall. Then the kinetic energy would be

$$K=\frac{1}{2}\int\limits_0^L\frac{m}{L}v(x)^2\,dx$$

If we suppose the uniform compression, we can say that the velocity is proportional to the current position along the spring

$$v(x)=\frac{x}{L}\dot L$$

Where $$\dot L=\frac{dL}{dt}$$. As such, we compute this integral and get

$$K=\frac{1}{6}m\dot L ^2$$

Looking at these expressions, this system behaves like a Simple Harmonic Oscillator (SHO) with mass $$m/3$$ and spring constant $$k$$. It should oscillate about $$L_0$$ with period

$$T=2\pi\sqrt\frac{m}{3k}$$

Which goes to 0 as $$m\rightarrow0$$. We then have oscillations of finite amplitude with zero period. The spring would oscillate like crazy at arbitrarily high speeds, which is clearly unphysical.

So, you can't neglect the mass of the spring when it is the only mass of the entire system.

While it is true that there isn't any mass attached to the spring, a real spring has mass itself, so after contracting to the natural length, the mass of the spring carries momentum. Thus, it will continue to move, resulting in the well-known oscillations, until the energy is completely dissipated by internal friction.

Edit: I am not sure if the model "massless spring without attached mass" does make any sense. There is no mass in the system that any force could act on.

• A "massless spring without attached mass" makes good sense if you are doing statics, not dynamics. It also makes sense if there is also a (massless) damping element in the dynamic model, but the equation of motion is then first order not second order. – alephzero Nov 30 '18 at 11:52
• I agree, but in a sense, by looking at the static case (i.e. stretching it, but not releasing it) you do attach a mass to it. Same for the massless damping element. – scaphys Nov 30 '18 at 12:18