GREPhysics.NET: GR0177 #92

GR0177 #92

Problem

GREPhysics.NET Official Solution

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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

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.

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uhurulol
2014-10-21 21:55:52

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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.)

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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!

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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).

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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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