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Saturday, 6 August 2011

Physics GRE - #5

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 :

Let the length of the stretched spring be x=x_1-x_2. From Hooke's Law we know that: m\ddot{x_{1}}=-k(x-l_0) m\ddot{x_{2}}=-k(x-l_0) Subtracting the two equations above, we obtain: m(\ddot{x_{1}}-\ddot{x_{2}})=m\ddot{x}=-\frac{2k}{m}x Thus the solution is: \omega_{0}^{2}=\frac{2k}{m}\implies\omega_{0}=\sqrt{\frac{2k}{m}}

3 comments:

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[hide] Alemi said...

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.

on 10 August 2011 at 10:57
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[hide] Paul Liu said...

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

on 10 August 2011 at 11:55
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[hide] Paul Liu said...

Where x is the displacement from equilibrium of course.

on 10 August 2011 at 11:56
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