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  GR8677 #35
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Verbatim question for GR8677 #35
Mechanics}Hamiltonian

Recall the lovely relation,

where T is the kinetic energy and V is the potential energy. (And, while one is at this, one should also recall that the Lagrangian is L=T-V.)

The potential energy is given to be kx^4. The kinetic energy \frac{1}{2}mv^2 can be written in the form of \frac{p^2}{2m}. Thus, choice (A) is the right answer.

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Comments
Almno10
2010-11-11 19:34:47
I always think of that skater scott hamilton. He used to do back flips... had tons of energy. Hamiltonian is total energy.

Lagrangian on the other hand? Lagrange multipliers are for lazy people. Hard workers would optimize functions by guess and check (lazy = less energy T-V).

whatever.
mpdude8
2012-04-15 21:05:06
Hey, man, whatever works. I have some weird ways of remembering random crap too.
NEC
a19grey2
2008-11-03 21:24:52
Also, if you are unsure about the sign of the kx^4 term you can just note that answer B) and D) are exactly equivalent through the relation noted by yosun in the original solution. Thus, ETS would never give TWO right answers and so the correct sign must be the unique answer given in A.

BUT... you really should remember the sign conventions for the Hamiltonian and the Lagrangian.
BerkeleyEric
2010-01-12 21:48:15
Well, I suppose if they wanted to be really tricky they could have also had an option for (1/2)mv^2+ kx^4 to test if you know that Hamiltonians are in phase space (x, p_x) and Lagrangians are in state space (x, dx/dt).
neon37
2010-11-01 10:46:35
@BerkeleyEric - that is true but that wouldnt make either of them wrong.

Important thing is to remember that Hamiltonian is total energy of the system where as Lagrangian is not.
NEC
Mexicorn
2005-11-08 12:11:38
Like the last problem, the potential should be U=kx^4
yosun
2005-11-09 02:24:56
Mexicorn: thanks for the typo-alert; it has been corrected.
Fixed Typos!

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