The Lagrangian of two coupled oscillators of mass m each is L = 12( dot x_1^2 + dot x_2^2 ) - 12m _0^2( x_1^2 + x_2^2 ) + m _0^2 x_1x_2 The equations of motion are

ddot x_1 + _0^2x_1 = _0^2 x_2,ddot x_2 + _0^2x_2 = _0^2 x_1

This question belongs to Engineering Physics Classical Mechanics

Classical Mechanics

The Lagrangian of two coupled oscillators of mass m each is $$L = \frac{1}{2}\left( {{{\dot x}_1}^2 + {{\dot x}_2}^2} \right) - \frac{1}{2}m{\omega _0}^2\left( {{x_1}^2 + {x_2}^2} \right) + m{\omega _0}^2\mu {x_1}{x_2}$$
The equations of motion are

A) $${{\ddot x}_1} + {\omega _0}^2{x_1} = {\omega _0}^2\mu {x_1},\,{{\ddot x}_2} + {\omega _0}^2{x_2} = {\omega _0}^2\mu {x_2}$$ B) $${{\ddot x}_1} + {\omega _0}^2{x_1} = {\omega _0}^2\mu {x_2},\,{{\ddot x}_2} + {\omega _0}^2{x_2} = {\omega _0}^2\mu {x_1}$$ C) $${{\ddot x}_1} + {\omega _0}^2{x_1} = {\omega _0}^2\mu {x_1},\,{{\ddot x}_2} + {\omega _0}^2{x_2} = - {\omega _0}^2\mu {x_2}$$ D) $${{\ddot x}_1} + {\omega _0}^2{x_1} = {\omega _0}^2\mu {x_1},\,{{\ddot x}_2} + {\omega _0}^2{x_2} = {\omega _0}^2\mu {x_1}$$

Answer & Solution

Correct Answer: Option B

Let's discuss the solution. Join the discussion

Question posted by Examveda

Community

Join the Discussion

Join the discussion

No comments yet Be the first to discuss this question.

Keep Learning

The Lagrangian of two coupled oscillators of mass m each is...

The Lagrangian of two coupled oscillators of mass m each is $$L = \frac{1}{2}\left( {{{\dot x}_1}^2 + {{\dot x}_2}^2} \right) - \frac{1}{2}m{\omega _0}^2\left( {{x_1}^2 + {x_2}^2} \right) + m{\omega _0}^2\mu {x_1}{x_2}$$ The equations of motion are

Join the Discussion

Related Questions on Classical Mechanics

The Lagrangian of two coupled oscillators of mass m each is $$L = \frac{1}{2}\left( {{{\dot x}_1}^2 + {{\dot x}_2}^2} \right) - \frac{1}{2}m{\omega _0}^2\left( {{x_1}^2 + {x_2}^2} \right) + m{\omega _0}^2\mu {x_1}{x_2}$$
The equations of motion are