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Mechanics}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).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
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}=0rnrn\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.Alternate Solution - Unverified
Comments
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
NEC
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}=0rnrn\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.Alternate Solution - Unverified

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