Physics Problems & Solutions: Classical Mechanics

The Lagrangian for a mechanical system is $L=a\dot q^2+bq^4$ where q is a generalized coordinate and a and b are constants. The equation of motion for this system is

$\\\\\\$A.$ \hspace{5 mm} \dot q=\sqrt{\frac{b}{a}}q^2\\\\$B.$ \hspace{5 mm} \dot q=\frac{2b}{a} q^3 \\\\$C.$ \hspace{5 mm}\ddot q=-\frac{2b}{a} q^3 \\\\$D.$ \hspace{5 mm} \ddot q=+\frac{2b}{a} q^3 \\\\$E.$ \hspace{5 mm} \ddot q=\frac{b}{a} q^3$

(GR0177 #74)

Solution:

The Lagrangian equation of motion for the generalized coordinate q:

$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}$

$\\\frac{\partial L}{\partial q}=\frac{\partial}{\partial q} (a\dot q^2+bq^4 )=0+\frac{\partial}{\partial q} bq^4=4bq^3\\\\\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial \dot q} (a\dot q^2+bq^4 )=\frac{\partial}{\partial \dot q} a\dot q^2+0=2a\dot q \\\\\\\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial t} 2a\dot q=2a \ddot q \\\\ \frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q} \rightarrow 4bq^3=2a\ddot q \rightarrow \ddot q=\frac{2b}{a} q^3$

Answer: D

Physics Problems & Solutions

Classical Mechanics - Lagrangian

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