$$\require{cancel}$$

11.2: Mass Attached to an Elastic Spring

I am thinking of a mass $$m$$ resting on a smooth horizontal table, rather than hanging downwards, because I want to avoid the unimportant distraction of the gravitational force (weight) acting on the mass. The mass is attached to one end of a spring of force constant $$k$$, the other end of the spring being fixed, and the motion is restricted to one dimension.

I suppose that the force required to stretch or compress the spring through a distance $$x$$ is proportional to $$x$$ and to no higher powers; that is, the spring obeys Hooke's Law. When the spring is stretched by an amount $$x$$ there is a tension $$kx$$ in the spring; when the spring is compressed by $$x$$ there is a thrust $$kx$$ in the spring. The constant k is the force constant of the spring.

When the spring is stretched by an distance $$x$$, its acceleration $$\ddot{x}$$ is given by

$m\ddot{x}=-kx. \label{11.2.1}$

This is an Equation of the type 11.1.5, with $$\omega^{2}=\frac{k}{m}$$, and the motion is therefore simple harmonic motion of period

$P=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{k}}. \label{11.2.2}$

At this stage you should ask yourself two things: Does this expression have dimensions T? Physically, would you expect the oscillations to be slow for a heavy mass and a weak spring? The reader might be interested to know (and this is literally true) that when I first typed Equation $$\ref{11.2.2}$$, I inadvertently typed $$\sqrt{\frac{k}{m}}$$ and I immediately spotted my mistake by automatically asking myself these two questions. The reader might also like to note that you can deduce that $$P_{\infty}\sqrt{\frac{m}{k}}$$ by the method of dimensions, although you cannot deduce the proportionality constant $$2\pi$$. Try it.

Energy Considerations. The work required to stretch (or compress) a Hooke's law spring by $$x$$ is $$\frac{1}{2}kx^{2}$$, and this can be described as the potential energy or the elastic energy stored in the spring. I shall not pause to derive this result here. It is probably already known by the reader, or s/he can derive it by calculus. Failing that, just consider that, in stretching the spring, the force increases linearly from 0 to $$kx$$, so the average force used over the distance $$x$$ is $$\frac{1}{2}kx$$ and so the work done is $$\frac{1}{2}kx^{2}$$.

If we assume that, while the mass is oscillating, no mechanical energy is dissipated as heat, the total energy of the system at any time is the sum of the elastic energy $$\frac{1}{2}kx^{2}$$ stored in the spring and the kinetic energy $$\frac{1}{2}mv^{2}$$ of the mass. (I am assuming that the mass of the spring is negligible compared with $$m$$.)

Thus

$E=\frac{1}{2}kx^{2}+\frac{1}{2}mv^{2} \label{11.2.3}$

and there is a continual exchange of energy between elastic and kinetic. When the spring is fully extended, the kinetic energy is zero and the total energy is equal to the elastic energy then, $$\frac{1}{2}ka^{2}$$ when the spring is unstretched and uncompressed, the energy is entirely kinetic; the mass is then moving at its maximum speed $$a\omega$$ and the total energy is equal to the kinetic energy then, $$\frac{1}{2}ma^{2}\omega^{2}$$. Any of these expressions is equal to the total energy:

$E=\frac{1}{2}kx^{2}+\frac{1}{2}mv^{2}=\frac{1}{2}ka^{2}=\frac{1}{2}ma^{2}\omega^{2} \label{11.2.4}$