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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: $\omega_{0}^{2}=\frac{2k}{m}\implies\omega_{0}=\sqrt{\frac{2k}{m}}$

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