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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:
$\frac{d}{dt}{\theta }_1+{\mathrm{\Omega }}^2{\theta }_1=K({\theta }_2-{\theta }_1)$
$\frac{d}{dt}{\theta }_2+{\mathrm{\Omega }}^2{\theta }_2=-K({\theta }_2-{\theta }_1)$
where ${\mathrm{\Omega }}^2=\frac{g}{l},K=\frac{k}{m}$
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 $\mathrm{\Omega }$ and $\sqrt{{\mathrm{\Omega }}^2+2K}$
What's the logic behind these equations? Widener (talk) 18:22, 27 May 2012 (UTC)

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)