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## Sunday, 14 August 2011

### Physics GRE - #13

Two pendula are attached to a massless spring, as shown below. The arms of the pendula are both of length $l$, but the pendulum balls have unequal masses $m_1$ and $m_2$. The initial distance between the masses is the equilibrium length of the spring, which has spring constant $K$. What is the highest normal mode frequency of this system?

1. $\sqrt{\frac{g}{l}}$
2. $\sqrt{\frac{K}{m_1+m_2}}$
3. $\sqrt{\frac{K}{m_1}+\frac{K}{m_2}}$
4. $\sqrt{\frac{g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$
5. $\sqrt{\frac{2g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$

Solution :

The correct answer is choice 4.

The easy way of solving this problem is to look at the limiting behaviour of the system in several ways. First we see that if $m_1\rightarrow\infty$ and $l\rightarrow\infty$, the resulting behaviour should be the same as a single mass, $m_2$, attached to a wall by a spring, thus we should have $\sqrt{\frac{K}{m_2}}$. This just leaves options 3 and 4 as possibilities. Then we note that if $K\rightarrow 0$, it would be as if there was no spring between the two pendula, and so we should get $\sqrt{\frac{g}{l}}$. This leaves choice 4 as the right answer. This solution sure beats deriving the normal frequencies from scratch (which I'm not even sure I remember how to do).

A note on looking at limiting behaviours
Looking at limiting behaviour is a useful technique for doing a quick check of your answer or in multiple choice questions such as these. Sometimes looking at a system this way can also give you nice approximate answers. Common methods of looking at limiting behaviour usually involves making all masses/lengths/constants equal in a problem, or taking them to 0/∞.

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