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Captain Hooke

  

Figure: Sketch of a mass on a spring. In the leftmost frame the mass m is at rest and the spring is in its equilibrium position ( i.e. neither stretched nor compressed). [If gravity is pulling the mass down, then in the equilibrium position the spring is stretched just enough to counteract the force of gravity. The equilibrium position can still be taken to define the  x=0  position.] 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.

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 (using vector notation to account for the directionality)

$$\mbox{\boldmath$\vec{F}$\unboldmath } = - k \, \mbox{\boldmath$\vec{x}$\unboldmath }$$ (11.11)

which is the mathematical expression of the concept of a linear restoring force. This is known as HOOKE'S LAW. It is a lot more general than it looks, as we shall see later.

Keeping in mind that the   $\mbox{\boldmath$\vec{F}$\unboldmath }$  given above is the force exerted by the spring against anyone or anything trying to stretch or compress it. If you are that stretcher/compressor, the force you exert is   $-\mbox{\boldmath$\vec{F}$\unboldmath }$. If you do work on the spring^11.9 by stretching or compressing it^11.10 by a differential displacement   $d\mbox{\boldmath$\vec{x}$\unboldmath }$  from equilibrium, the differential amount of work done is given by

$$dW \; = \; - \mbox{\boldmath $\vec{F}$\unboldmath } \cdot d . . . . . . d\mbox{\boldmath $\vec{x}$\unboldmath } \; = \; k \, x \, dx$$

which we can integrate from  x = 0  (the equilibrium position) to  x  (the final position) to get the net work  W:

$$W = k \int_0^x x \, dx = {1\over2} k \, x^2$$ (11.12)

Once you let go, the spring will do the same amount of work back against the only thing trying to impede it - namely, the inertia of the mass  m  attached to it. This can be used with the WORK AND ENERGY Law to calculate the speed   $v_{\rm max}$  in the third frame of Fig. 11.4: since  v₀ = 0,

$${1\over2} m \, v_{\rm max}^2 = {1\over2} k \, x_{\rm max}^2 . . . . . . {\rm or} \qquad v_{\rm max}^2 = {k \over m} \; x_{\rm max}^2$$

$$\hbox{\rm or} \qquad v_{\rm max} = \sqrt{k \over m} \; \; \vert x_{\rm max}\vert$$

where   $\vert x_{\rm max}\vert$  denotes the absolute value of   $x_{\rm max}$  (i.e. its magnitude, always positive). Note that this is a relationship between the maximum values of  v  and  x, which occur at different times during the process.