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The Spring Pendulum

Another mecahnical example will serve to establish the paradigm of SIMPLE HARMONIC MOTION (SHM) as a solution to a particular type of equation of motion.13.4

Figure: Successive "snapshots" of a mass bouncing up and down on a spring.
\epsfig{file=PS/,height=3.15in} }\end{center}\end{figure}

As discussed in a previous chapter, the spring embodies one of Physics' premiere paradigms, the linear restoring force. That is, a force which disappears when the system in question is in its "equilibrium position"  x0  [which we will define as the  x = 0  position   $(x_0 \equiv 0)$  to make the calculations easier] but increases as  x  moves away from equilibrium, in such a way that the magnitude of the force  F  is proportional to the displacement from equilibrium [F  is linear in  x] and the direction of  F  is such as to try to restore  x  to the original position. The constant of proportionality is called the spring constant, always written  k. Thus  F = - k x  and the resultant equation of motion is

\begin{displaymath}\ddot{x} = - \left( k \over m \right) \, x
\end{displaymath} (13.10)

Note that the mass plays a rôle just as essential as the linear restoring force in this paradigm. If  $m \to 0$  in this equation, then the acceleration becomes infinite and in principle the spring would just return instantaneously to its equilibrium length and stay there!

In the leftmost frame of Fig. 13.4 the mass m is at rest and the spring is in its equilibrium position (i.e. neither stretched nor compressed) defined as  x=0.  In the second frame, the spring has been gradually pulled down a distance   $x_{\rm max}$  and the mass is once again at rest. Then the mass is released and accelerates upward under the influence of the spring until it reaches the equilibrium position again [third frame]. This time, however, it is moving at its maximum velocity   $v_{\rm max}$  as it crosses the centre position; as soon as it goes higher, it compresses the spring and begins to be decelerated by a linear restoring force in the opposite direction. Eventually, when   $x = -x_{\rm max}$, all the kinetic energy has been been stored back up in the compression of the spring and the mass is once again instantaneously at rest [fourth frame]. It immediately starts moving downward again at maximum acceleration and heads back toward its starting point. In the absence of friction, this cycle will repeat forever.

I now want to call your attention to the acute similarity between the above differential equation and the one we solved for exponential decay:

\begin{displaymath}\dot{x} = - \kappa \, x
\end{displaymath} (13.11)

and, by extension,

\begin{displaymath}\ddot{x} = \kappa^2 \, x
\end{displaymath} (13.12)

the solution to which equation of motion (i.e. the function which "satisfies" the differential equation) was

\begin{displaymath}x(t) = x_0 \, e^{- \kappa \, t}
\end{displaymath} (13.13)

Now, if only we could equate  $\kappa^2$  with  - k/m,  these equations of motion (and therefore their solutions) would be exactly the same! The problem is, of course, that both  k  and  m  are intrinsically positive constants, so it is tough to find a real constant  $\kappa$  which gives a negative number when squared.

next up previous
Next: Imaginary Exponents Up: Simple Harmonic Motion Previous: Simple Harmonic Motion
Jess H. Brewer - Last modified: Sun Nov 15 13:36:04 PST 2015