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GR0177 #74
Problem
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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
 insertphyspun2011-02-21 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
 antonis2013-09-14 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 insertphyspun2011-02-21 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

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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