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: Option B
This Question Belongs to Engineering Physics >> Classical Mechanics
A. increases till mass falls into hole
B. decreases till mass falls into hole
C. remains constant
D. becomes zero at radius r1, where 0 < r1 < r0
A. $$\frac{c}{3}$$
B. $$\frac{{\sqrt 2 }}{3}c$$
C. $$\frac{c}{2}$$
D. $$\frac{{\sqrt 3 }}{2}c$$
The Hamiltonian corresponding to the Lagrangian $$L = a{{\dot x}^2} + b{{\dot y}^2} - kxy$$ is
A. $$\frac{{{p_x}^2}}{{2a}} + \frac{{{p_y}^2}}{{2b}} + kxy$$
B. $$\frac{{{p_x}^2}}{{4a}} + \frac{{{p_y}^2}}{{4b}} - kxy$$
C. $$\frac{{{p_x}^2}}{{4a}} + \frac{{{p_y}^2}}{{4b}} + kxy$$
D. $$\frac{{{p_x}^2 + {p_y}^2}}{{4ab}} + kxy$$
A. circular
B. elliptical
C. parabolic
D. hyperbolic
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