GREPhysics.NET: GR0177 #74

GR0177 #74

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }Lagrangians

The Lagrangian equation of motion is given by $\frac{\partial L}{\partial q} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ for the generalized coordinate q.

Chunking out the derivatives, one finds that

$\frac{\partial L}{\partial q}=4bq^3$

$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=2a\ddot{q}$

Setting the two equal to each other as in the Lagrangian equations of motion given above (without undetermined multipliers), one finds that $2bq^3=a\ddot{q}$ , which gives choice (D).

Alternate Solutions

insertphyspun
2011-02-21 14:39:38

Say one does not remember the Lagrangian equation of motion (maybe you should memorize it now). One can still get the answer by conservation of energy.rnrn $\frac{dE}{dt}=0$ rnrn $\frac{dE}{dt}=2a\dot{q}\ddot{q}-4bq^3\dot{q}$ rnrnSolve for $\ddot{q}$ and find $\ddot{q}=2\frac{b}{a}q^3$ . Answer D.

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antonis 2013-09-14 09:29:15	Just another way - I think - recalling that the Lagrangian usually takes the form $L=T-V$ . This Lagrangian could bring us in mind a body with kinetic energy $T=\frac{1}{2}m\dot{q}^2=a \dot{q}^2$ in a potential well $V=-bq^4$ , meaning that the force exerted to the body would be $F=-\frac{\partial V}{\partial q}=4bq^3} \Rightarrow \ddot{q}=\frac{4bq^3}{m}=\frac{2b}{a}q^3$ Reply to this comment
insertphyspun 2011-02-21 14:39:38	Say one does not remember the Lagrangian equation of motion (maybe you should memorize it now). One can still get the answer by conservation of energy.rnrn $\frac{dE}{dt}=0$ rnrn $\frac{dE}{dt}=2a\dot{q}\ddot{q}-4bq^3\dot{q}$ rnrnSolve for $\ddot{q}$ and find $\ddot{q}=2\frac{b}{a}q^3$ . Answer D. Reply to this comment

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Just another way - I think - recalling that the Lagrangian usually takes the form $L=T-V$ . This Lagrangian could bring us in mind a body with kinetic energy $T=\frac{1}{2}m\dot{q}^2=a \dot{q}^2$ in a potential well $V=-bq^4$ , meaning that the force exerted to the body would be $F=-\frac{\partial V}{\partial q}=4bq^3} \Rightarrow \ddot{q}=\frac{4bq^3}{m}=\frac{2b}{a}q^3$

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