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GR0177 #92
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Mechanics$\Rightarrow$}Hamiltonian

The Hamiltonian is just the sum of the kinetic and potential energy, $H=T+V$.

The kinetic energy due to each mass is $T_i = p_i^2/(2m)$. The potential energy is just $U=1/2 k (\Delta l)^2$. $\Delta l = l -l_0$, and thus factoring out the $1/2$, one arrives at choice (E).

Alternate Solutions
 uhurulol2014-10-21 21:55:52 Process of elimination solution: Hamiltonian $\Rightarrow$ $H = T + V$, a minus in front of the potential $V$ would indicate a Lagrangian. Eliminate (A) (C) and (D). Spring potential energy and kinetic energy are both proportional to $\frac{1}{2}$, not $1$ or $2$, eliminate (B). All that's left is (E). Hooray.Reply to this comment
uhurulol
2014-10-21 21:55:52
Process of elimination solution:

Hamiltonian $\Rightarrow$ $H = T + V$, a minus in front of the potential $V$ would indicate a Lagrangian. Eliminate (A) (C) and (D).

Spring potential energy and kinetic energy are both proportional to $\frac{1}{2}$, not $1$ or $2$, eliminate (B).

All that's left is (E). Hooray.
Barney
2012-11-07 09:18:00
The Test-Takers Solution:
The Hamiltonian is the Energy E of the system - pick choice (E).

(Funny to see that ETS people seem to like little eastereggs. This is not the only case where you can observe such "accidental" matching.)
jmason86
2009-07-05 20:04:03
If you didn't know the potential energy of the spring, you could derive it from Hooke's law.

$F = -kx$
In general, $U=\int F$, so
$U = -\frac{1}{2}kx^2$

What I don't get is that the negative in Hooke's law doesn't just disappear because you integrate.. so why is potential energy stored in a spring defined to be positive? This would lead to the incorrect answer of (D).
 jmg8102009-07-08 14:19:25 If a force is derivable from a position dependent potential$\,U$ then $\vec{F}$ = -$\nabla{U}$($\vec{r}$), and hence $\,U = -\int{F}\,d\vec{r}$.
 jmg8102009-07-09 09:20:29 Typo fix: $\,U = -\int\vec{F}\cdot\,d\vec{r}$
 jmason862009-07-23 17:36:11 Thanks!
neutrino
2007-10-31 03:34:57
I don't understand the signs. If the string is fully stretched", $T=0$, but $V$ is max. I would guess answer D in this case.

What is my error?
 hot_dark_matter2008-05-23 15:00:49 If the spring is fully stretched, it doesn't mean that the kinetic energy is zero. Presumably, the directions of $\mathbf{p}_1$ and $\mathbf{p}_2$ indicate the system can have translational and rotational energy. As for the signs, kinetic energy is always positive and so is the potential energy stored by a spring. This eliminates all choices but (B) and (E). Given that there is only one spring, the answer must be (E).

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