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Friday, 12 August 2011

Physics GRE - #11

A pendulum of length $l$ is attached to the roof of an elevator near the surface of the Earth.

The elevator moves upward with acceleration $\frac{1}{2}g$. The angular frequency is:

  • $\sqrt{\frac{3g}{2l}}$
  • $\sqrt{\frac{2g}{3l}}$
  • $\sqrt{\frac{g}{l}}$
  • $\sqrt{\frac{g}{2l}}$
  • $\sqrt{\frac{2g}{l}}$

Solution :

Quick solution:
The angular frequency of a pendulum is $\omega=\sqrt{\frac{g}{l}}$ when the gravitational field has value $g$. In this problem, the effective gravity is $g_{eff} = g + \frac{1}{2}g = \frac{3}{2}g$. \[\therefore \omega=\sqrt{\frac{g_{eff}}{l}}=\sqrt{\frac{3g}{2l}}.\]

Full solution: 
By Newton's 2nd law for rotational motion, \[\sum{\tau}=I\alpha \implies -mg_{eff}l\sin\theta=I\ddot{\theta}.\]
Substituting $I = ml^2$ and $g_{eff} = \frac{3}{2}g$, we have
\[-\frac{3}{2}mgl\sin\theta=ml^2\ddot{\theta}\implies \ddot{\theta}=-\frac{3g}{2l}\sin\theta\approx-\frac{3g}{2l}\theta\] \[\therefore \omega^2 = \frac{3g}{2l} \implies \omega=\sqrt{\frac{3g}{2l}}.\]


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