- #1

Nathanael

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## Homework Statement

A system consisting of a rod of mass M and length L is pivoted at its centre P. Two springs of spring constants k

_{1}and k

_{2}are attached as shown. They are relaxed when the rod is horizontal. What is the time period of the rod if it is given a slight angular displacement.

## Homework Equations

M = 200

k

_{1}= 20

k

_{2}= 2

Presumably all in SI units

## The Attempt at a Solution

I would describe the situation by the equation

[itex](\frac{mL^2}{12})\frac{d^2\theta}{dt^2}=-\frac{L^2}{4}(k_1+k_2)\theta[/itex]

Because [itex]\frac{mL^2}{12}[/itex] is the rotational inertia, and [itex]\frac{L^2}{4}(k_1+k_2)\theta[/itex] is (approximately) the torque.

We don't care about the angle at time zero, so we don't need the most general solution (all solutions should have the same period) we just need a particular solution, which could be:

[itex]\theta_{max}\sin(\sqrt{\frac{3(k_1+k_2)}{m}}t)=\theta_{max}\sin(\sqrt{\frac{66}{200}}t)[/itex]

The period should then be [itex]T=2\pi \sqrt{\frac{100}{33}}\approx 10.9[/itex] seconds

But the answer is apparently 6.62 seconds