Hooke's law:

$\overline{){{\mathbf{F}}}_{\mathbf{s}\mathbf{p}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}}{\mathbf{=}}{\mathbf{-}}{\mathbf{k}}{\mathbf{x}}}$

Both the springs will experience a force, F.

For spring 1, we'll have:

$\mathit{F}\mathbf{=}{\mathit{k}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}$

$\mathrm{T}=2\mathrm{\pi}\sqrt{\frac{\mathrm{M}}{\mathrm{k}}}$

The period of an object oscillating on the end of a spring is given by the formula:

Here, the spring constant k comes from the usual definition by Hooke’s law, in terms of the force of the spring at some displacement from the equilibrium position x1:

F = -k (x-x_{0})

If we let M denote the mass of the object oscillating, this accurately describes an ideal situation in which springs are massless. There is actually a correction when we consider the case of springs whose mass of the springs, then the effective mass M (which we plug into the above formula) is defined by:

M = M_{G} + M_{s}/3

We will also make use of the potential energy stored in a spring:

U = ½ k (x-x0)^{2}

Note that the kinetic energy also uses the same effective mass M in place of the mass of the glider.

Effective Spring Constant: In Part 1, you measured the k_{effective} fo the two springs acting together. If the two springs k1 and k2 individually, how would they combine to get k_{eff}?

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