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GR8677 #35
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Mechanics$\Rightarrow$}Hamiltonian

Recall the lovely relation,

$H=T+V,
$

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.

Alternate Solutions
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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.
 mpdude82012-04-15 21:05:06 Hey, man, whatever works. I have some weird ways of remembering random crap too.
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.
 BerkeleyEric2010-01-12 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).
 neon372010-11-01 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.
Mexicorn
2005-11-08 12:11:38
Like the last problem, the potential should be $U=kx^4$
 yosun2005-11-09 02:24:56 Mexicorn: thanks for the typo-alert; it has been corrected.

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