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Tuesday, 9 August 2011

Physics GRE - #8

A particle of mass $m$ moves in the potential shown above. The period of the motion when the particle has energy $E$ is:
  • $\sqrt{\frac{k}{m}}$
  • $2\pi\sqrt{\frac{m}{k}}$
  • $2\sqrt{\frac{2E}{mg^2}}$
  • $\pi\sqrt{\frac{m}{k}}+2\sqrt{\frac{2E}{mg^2}}$
  • $2\pi\sqrt{\frac{m}{k}}+4\sqrt{\frac{2E}{mg^2}}$

Solution :

The potential function $V(x)$ is half of a simple harmonic oscillator potential and a full gravitational potential. To find the period, we can simply add half the period of a simple harmonic oscillator to twice the period of a falling object (since we know there are 2 parts, we know that the answer is one of the last two choices at this point).

Recall that the period of a harmonic oscillator is $T=2\pi\sqrt{\frac{m}{k}}$. Half of this period is $T_1=\pi\sqrt{\frac{m}{k}}$. From this we know that the second last choice must be the answer as the last choice has a factor of 2 in front of it. To check our answer, we can find the period of a falling object from the kinematics equation: $x=\frac{1}{2}gt^2$, giving us $T=\sqrt{\frac{2x}{g}}$. At the endpoints of the oscillation, we have $v=0$ so $mgx=E$. Substituting for $x$ in $T$  as well as taking into account of the fact that the particle has to go to the endpoint and back (this introduces a factor of 2), the period contribution from the gravitational potential is $T_2=2\sqrt{\frac{2E}{mg^2}}$. Thus the period is: \[T_{total}=T_1+T_2=\pi\sqrt{\frac{m}{k}}+2\sqrt{\frac{2E}{mg^2}}\] as we suspected.


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