GR9677 #84



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Comments 
a 20191009 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 http://www.gbservices.net/tobuyen10.asp   enterprise 20180407 14:27:53  There is another way to work this out. Any twobody 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.
enterprise 20180407 14:28:52 
Sorry. It appears in B also , But this doesn\'t involve g/l

  whereami 20180406 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 unphysical. The only way you see m1+m2 is when you put m2 and m1 together on the same side of a spring.   insertphyspun 20110221 11:45:45  I didn't think to take , which I think is the superior way to do this problem. But I did take and asked myself what the effective spring constant of the system would be if . Consider the following argument:
If , each mass moves in the highest normal mode, which is equivalent to one mass moving . The potential energy of the system is therefore , giving .
The frequency of oscillation should then be .
Answer (D) has the correct frequency.   gman 20101111 13:45:23  Take , should get something that depends on since is still attached by the spring. Eliminate A, B and E.
Now in that limit, take , which is a regular pendulum. Should reduce to . Eliminate C.
Pick D.   fjornada 20091018 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 20100730 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).

abcdefghijk 20180405 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

  hungrychemist 20071008 22:47:17  take limit as K> 0, you should get the lowest normal mode frequency
resinoth 20150918 23:06:29 
nice.

  grep 20061027 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 20070802 20:10:18 
I don't think dimensional analysis can be used here. It certainly does not eliminate B or E.

wizang 20091002 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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