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11.2: Mass Attached to an Elastic Spring

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    6996
  • [ "article:topic", "authorname:tatumj" ]

    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:

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

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