GREPhysics.NET: GR8677 #35

GR8677 #35

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Mechanics $\Rightarrow$ }Hamiltonian

Recall the lovely relation,

$H=T+V,<br />$

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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Comments

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).

Reply to this comment

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.

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