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May 27

Coupled pendulurms

Two identical pendulums of length $l$ are coupled by means of a spring of natural length $a$ and spring constant $k$ . When the displacement angles $\theta _{1}(t)$ and $\theta _{2}(t)$ are small, the equations of motion are:
${\frac {d}{dt}}\theta _{1}+\Omega ^{2}\theta _{1}=K(\theta _{2}-\theta _{1})$
${\frac {d}{dt}}\theta _{2}+\Omega ^{2}\theta _{2}=-K(\theta _{2}-\theta _{1})$
where $\Omega ^{2}={\frac {g}{l}},K={\frac {k}{m}}$
Here $m$ is the mass of the bendulum bob and $g$ is the acceleration due to gravity.
Find the frequencies of the normal modes of oscillation. Give a simple justification of the above equations (a full derivation is not requried).

I'm not sure about the normal modes of oscillation; I get $\Omega$ and ${\sqrt {\Omega ^{2}+2K}}$
What's the logic behind these equations? Widener (talk) 18:22, 27 May 2012 (UTC)[reply]

They are just derived from considering the forces on the pendulum weights. Multiply through by m and by l and you have the force due to the displacement (assuming it's small) and spring on each mass and mass × acceleration.--JohnBlackburne^words_deeds 00:12, 28 May 2012 (UTC)[reply]