## Wednesday, September 28, 2011

### Lagrangian problem

Question 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. $\dot q=\sqrt{\frac{b}{a}}q^2$
B. $\dot q=\frac{2b}{a} q^3$
C. $\ddot q=\frac{-2b}{a} q^3$
D. $\ddot q=+\frac{2b}{a} q^3$
E. $\ddot q=\frac{b}{a} q^3$
Solution:
The Lagrangian equation of motion for the generalized coordinate $q$:
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}$
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial q} (a\dot q^2+bq^4 )=0+\frac{\partial}{\partial q} bq^4=4bq^3$
$\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial \dot q} (a\dot q^2+bq^4 )=\frac{\partial}{\partial \dot q} a\dot q^2+0=2a\dot q$
$\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial t} 2a\dot q=2a \ddot q$
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q} \rightarrow 4bq^3=2a\ddot q \rightarrow \ddot q=\frac{2b}{a} q^3$
source :-http://physics-problems-solutions.blogspot.com