Hooke's Law

Hooke's Law says that the restoring force due to a spring is proportional to the length that the spring is stretched, and acts in the opposite direction. If we imagine that there are no other forces, and let x(t) represent the distance the spring is stretched at time t then the restoring force might be represented as -K²x(t) where K > 0.

On the other hand, we accept the relation that the total force on an object is the product of its acceleration and its mass (our old friend I. Newton). Since acceleration, a, is the rate of change of velocity, v, with respect to time, and velocity is the rate of change of position with respect to time, we have a(t) = v'(t) and v(t) = x'(t), so a(t) = (x')'(t), that is, acceleration is the derivative of the derivative of position with respect to time. Usually we drop the parentheses and write x'' in place of (x')', and refer to x'' as the second derivative of x (with respect to whatever the variable is). Thus for our spring, the total force is mx''(t), where m > 0 s the mass of the spring. Thus Hooke's Law tells us that if there are no other forces (no gravity, no air resistance, etc.) then

mx''(t) = -K²x(t)

Since m > 0 we may write this as

$$x''(t) = - \left(\frac{K}{\sqrt{m}}\right)^2x(t).$$

If the mass and the spring constant were equal to 1, we would immediately recognize that we want functions whose second derivatives are their opposites. We know two such functions, sine and cosine. By considering what happens when we differentiate f(cx) with respect to x, we conclude that a function x whose rule is

$$x(t) = A\cos\left(\frac{K}{\sqrt{m}}t\right) + B\sin\left(\frac{K}{\sqrt{m}}t\right)$$

will always satisfy () if A and B are constants. Which A and B do we choose? Well, we haven't specified everything about the spring, as we haven't said how far it was initially stretched, and how fast it was moving at the start. That is, we haven't specified x(0) and x'(0). These determine A and B since

$$\begin{array} {rcl} x(0) & = & A\cos(0) + B\sin(0)\\ & = & A... ...t{m}}} \\ & = & {\displaystyle B\frac{K}{\sqrt{m}} }\end{array}$$

Again, we need to satisfy ourselves that no other solution is possible, and we will do that a little later on.