Three particles of mass m each situated at x₁(t), x₂(t) and x₃(t) respectively are connected by two spring constants k and unstretched lengths $$l$$. The system is free to oscillate only in one-dimension along the straight line joining all the three particles. The Lagrangian of the system is

A. $$L = \frac{m}{2}\left[ {{{\left( {\frac{{d{x_1}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_2}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_3}}}{{dt}}} \right)}^2}} \right] - \frac{k}{2}{\left( {{x_1} - {x_2} - l} \right)^2} + \frac{k}{2}\left( {{x_3} - {x_2} - l} \right)$$

B. $$L = \frac{m}{2}\left[ {{{\left( {\frac{{d{x_1}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_2}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_3}}}{{dt}}} \right)}^2}} \right] - \frac{k}{2}{\left( {{x_1} - {x_3} - l} \right)^2} + \frac{k}{2}\left( {{x_3} - {x_2} - l} \right)$$

C. $$L = \frac{m}{2}\left[ {{{\left( {\frac{{d{x_1}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_2}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_3}}}{{dt}}} \right)}^2}} \right] - \frac{k}{2}{\left( {{x_1} - {x_2} - l} \right)^2} - \frac{k}{2}{\left( {{x_3} - {x_2} + l} \right)^2}$$

D. $$L = \frac{m}{2}\left[ {{{\left( {\frac{{d{x_1}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_2}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_3}}}{{dt}}} \right)}^2}} \right] - \frac{k}{2}{\left( {{x_1} - {x_2} - L} \right)^2} - \frac{k}{2}{\left( {{x_3} - {x_2} - l} \right)^2}$$

Answer: Option A