Two equal masses $m_1=m_2=m$ are connected by a spring having Hooke's constant k. If the equilibrium separation is $l_0$ and the spring rests on a frictionless horizontal surface, then the angular frequency $\omega_0$ is:

- $\sqrt{\frac{k}{m}}$
- $\sqrt{\frac{2k}{m}}$
- $\sqrt{\frac{3k}{m}}$
- $2\sqrt{\frac{k}{m}}$
- $\sqrt{\frac{g}{l_0}}$

Solution :

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omega^2 = ( restoring force ) / ( (unit displacement) * (unit mass) )

No longer a need to memorize the frequency of a 12 different systems, this works for them all: pendulums, circuits, even ones you've never seen before like a hoop on a nail.

I always remember it as $\ddot{x}=-\omega^2 x$, but I guess it's the same thing.

Where $x$ is the displacement from equilibrium of course.

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