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GR9677 #84
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Alternate Solutions |
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Comments |
fjornada
2009-10-18 19:56:52 | take  and  .
that's just the case of a mass ( ) 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).
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|  | 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
welshmj
2007-08-02 20:10:18 |
I don't think dimensional analysis can be used here. It certainly does not eliminate B or E.
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wizang
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 
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