Discussion
If the Lagrangian \(L(x, \dot{x})\) corresponding to Atwood's machine is given as :
\[ \begin{aligned} L(x, \dot{x}) &= \tfrac{1}{2}(m_{1} + m_{2})\dot{x}^{2} \\ &\quad + m_{1}gx + m_{2}g(l - x), \end{aligned} \]
where \(m_{1}\), \(m_{2}\), \(l\) and \(g\) are constants. Then the equation of motion is :
Answer: C
The equation of motion is \(\ddot{x} = \dfrac{m_{1} - m_{2}}{m_{1} + m_{2}}g\).
Lagrange's equation is \(\tfrac{d}{dt}L_{\dot{x}} - L_{x} = 0\), where subscripts denote partial derivatives. Here \(\partial L/\partial \dot{x} = (m_{1} + m_{2})\dot{x}\) and \(\partial L/\partial x = m_{1}g - m_{2}g\), so
\[ (m_{1} + m_{2})\ddot{x} = (m_{1} - m_{2})g \]
The heavier mass accelerates down with this reduced gravity. An equation of motion gives the acceleration \(\ddot{x}\), so the two options with \(\dot{x}\) are wrong, and inverting the fraction would give an acceleration greater than \(g\), which is impossible.
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