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Mechanics}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).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
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.
Alternate Solution - Unverified
Comments
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.
Alternate Solution - Unverified
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.)
NEC
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).
jmg810
2009-07-08 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}.
jmg810
2009-07-09 09:20:29
Typo fix:

\,U = -\int\vec{F}\cdot\,d\vec{r}
jmason86
2009-07-23 17:36:11
Thanks!
NEC
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_matter
2008-05-23 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).
Answered Question!

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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.

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