- $\sqrt{\frac{g}{l}}$
- $\sqrt{\frac{K}{m_1+m_2}}$
- $\sqrt{\frac{K}{m_1}+\frac{K}{m_2}}$
- $\sqrt{\frac{g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$
- $\sqrt{\frac{2g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$

Solution :

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?

**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/∞.

- $\sqrt{\frac{g}{l}}$
- $\sqrt{\frac{K}{m_1+m_2}}$
- $\sqrt{\frac{K}{m_1}+\frac{K}{m_2}}$
- $\sqrt{\frac{g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$
- $\sqrt{\frac{2g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$

Solution :

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

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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