# Problem: T = 2πMkThe 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-x0)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 = MG + Ms/3We will also make use of the potential energy stored in a spring:U = ½ k (x-x0)2Note 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 keffective fo the two springs acting together. If the two springs k1 and k2 individually, how would they combine to get keff?

###### FREE Expert Solution

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:

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

84% (150 ratings) ###### Problem Details

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-x0)

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 = MG + Ms/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 keffective fo the two springs acting together. If the two springs k1 and k2 individually, how would they combine to get keff?

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Based on our data, we think this problem is relevant for Professor Gornea's class at Carleton University.