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Mechanics}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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2019-10-09 23:06:25
Puis, quand MMA est venu autour, je savais que ?a y était, que je pouvais faire les deux en même temps.\r\na
2018-04-07 14:27:53
There is another way to work this out. Any two-body problem in mechanics can be reduced to one body problem using the concept of reduced mass. So we should expect that the masses should appear as reduced mass in the solution. The only choice in which this happens is D which have K(1/M_1+1/M_2) =K/reduced mass. So , this is the solution.
2018-04-07 14:28:52
Sorry. It appears in B also , But this doesn\'t involve g/l
2018-04-06 03:32:09
i think the best way to solve this problem is some sense or knowledge of how the solution of coupled harmonic oscillators look like. Look at m1 and m2. They are on the opposite sides of a spring. This means: you will never see m1 +m2 as a factor in the solution. If you are familiar with some particular solutions of the coupled harmonic oscillator system, you will know this. This mathematical expression is simply un-physical. The only way you see m1+m2 is when you put m2 and m1 together on the same side of a spring. NEC
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.
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 dudeness
2010-07-30 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).
2018-04-05 03:13:54
well I got down to D and E as well. But instead of taking m to infinity, I took it to zero. In that way D gives an infinitely large frequency. I can\'t make sense out of it. I think a massless m2 simply makes itself disappeared so the system is just a pendulum with a meaningless spring attached it oscillating with f = sqrt (l/g). \r\nThere is something wrong with this reasoning. I hope somebody can help me eventually.\r\n
2007-10-08 22:47:17
take limit as K-> 0, you should get the lowest normal mode frequency
2015-09-18 23:06:29
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
2007-08-02 20:10:18
I don't think dimensional analysis can be used here. It certainly does not eliminate B or E.
2009-10-02 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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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.

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