Consider the Lagrangian $ L = m\dot{x}\dot{y} - mw_0^2 zxy $. If $ p_x $ and $ p_y $ denote the generalized momenta conjugate to $ x $ and $ y $, respectively, then the canonical equations of motion are _______.

Question

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The correct option is (C) :$\dot{x} = \frac{P_y}{m} $, $\dot{P_x} = -mw_0^2 y $, $\dot{y} = \frac{P_x}{m} $, $\dot{P_y} = -mw_0^2 x $

Answer

$\dot{x} = \frac{P_x}{m} $, $\dot{P_x} = -mw_0^2 x $, $\dot{y} = \frac{P_y}{m} $, $\dot{P_y} = -mw_0^2 y $

Answer

$\dot{x} = \frac{P_x}{m} $, $\dot{P_x} = mw_0^2 x $, $\dot{y} = \frac{P_y}{m} $, $\dot{P_y} = mw_0^2 y $

Answer

$\dot{x} = \frac{P_y}{m} $, $\dot{P_x} = -mw_0^2 y $, $\dot{y} = \frac{P_x}{m} $, $\dot{P_y} = -mw_0^2 x $

Answer

$\dot{x} = \frac{P_y}{m} $, $\dot{P_x} = mw_0^2 y $, $\dot{y} = \frac{P_x}{m} $, $\dot{P_y} = mw_0^2 x $

Consider the Lagrangian \( L = m\dot{x}\dot{y} - mw_0^2 zxy \). If \( p_x \) and \( p_y \) denote the generalized momenta conjugate to \( x \) and \( y \), respectively, then the canonical equations of motion are _______.

The Correct Option is C

Solution and Explanation

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