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.
A nice meta-formula to remember is that for all simple harmonic motion
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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