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GR9677 #84
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Mechanics$\Rightarrow$}Normal Mode

One can work through the formalism of the usual normal mode analysis or learn how to deal with normal mode frequencies the easy way:

The highest normal mode frequency is due to the two masses oscillating out of phase. The $\omega^2$ contribution from the pendulum is just $g/l$. The $\omega^2$ contribution from each mass due to the spring is $k_i/m_i$. This is choice (D).

(If one had to guess, one can immediately eliminate choice (A), since that is the lowest normal mode frequency. In normal modes, there's always an in-phase frequency, which tends to be the lowest frequency.)

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insertphyspun
2011-02-21 11:45:45
I didn't think to take $m_1\rightarrow\infty$, which I think is the superior way to do this problem. But I did take $l\rightarrow\infty$ and asked myself what the effective spring constant of the system would be if $m_1=m_2$. Consider the following argument:

If $m_1=m_2$, each mass moves $x$ in the highest normal mode, which is equivalent to one mass moving $2x$. The potential energy of the system is therefore $U=(1/2)k(2x)^2$, giving $k_{eff}=4k$.

The frequency of oscillation should then be $\omega=(4k/2m)^{1/2}=(2k/m)^{1/2}$.

Answer (D) has the correct frequency.
gman
2010-11-11 13:45:23
Take $m_{1} \rightarrow \infty$, should get something that depends on $K$ since $m_{2}$ is still attached by the spring. Eliminate A, B and E.

Now in that limit, take $K \rightarrow 0$, which is a regular pendulum. Should reduce to $\sqrt{g/l}$. Eliminate C.

Pick D.
2009-10-18 19:56:52
take $l \rightarrow \infty$ and $m_1 \rightarrow \infty$.
that's just the case of a mass ($m_2$) oscillating around a fixed point. only (D) has such a limit.
 his dudeness2010-07-30 05:30:04 Perhaps once you've got it down to choices (D) and (E), you can argue that if you let one of the masses go to infinity so that it's effectively fixed, the normal mode frequency should have some dependence on K, and a larger K will lead to a larger frequency. In (E), the K dependence drops out for this case, so that leaves choice (D).
hungrychemist
2007-10-08 22:47:17
take limit as K-> 0, you should get the lowest normal mode frequency
grep
2006-10-27 14:20:32
If normal mode oscillation isn't your thing, you can use dimensional analysis to eliminate B and E. Then just remember that in general for pendulum systems the frequency decreases when l increases. This leaves only choice D
 welshmj2007-08-02 20:10:18 I don't think dimensional analysis can be used here. It certainly does not eliminate B or E.
 wizang2009-10-02 00:53:46 I checked the units, and they work out on all of the possible answers. Keep in mind that the spring constant is measured in terms of force per unit length. Taking this into consideration, all the answers end up at $\frac{1}{s}$

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