 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/7/a/solution

For a zero-natural-length spring, extension is the full distance between its endpoints, so the squared spring lengths are quadratic functions of the Cartesian coordinates. The Lagrangian is

L = \frac{m _{1}}{2} (\dot{X}_{1}^{2} + \dot{Y}_{1}^{2}) + \frac{m _{2}}{2} (\dot{X}_{2}^{2} + \dot{Y}_{2}^{2}) - \frac{k _{1}}{2} (X_{1}^{2} + Y_{1}^{2}) - \frac{k _{2}}{2} [(X_{2} - X_{1})^{2} + (Y_{2} - Y_{1})^{2}] - g (m_{1} Y_{1} + m_{2} Y_{2}) .

The gravitational potential energy has positive derivative in the upward coordinate, giving downward forces. Applying the Euler-Lagrange equations to all four coordinates gives

m_{1} \ddot{X}_{1} + (k_{1} + k_{2}) X_{1} - k_{2} X_{2} m_{2} \ddot{X}_{2} + k_{2} X_{2} - k_{2} X_{1} m_{1} \ddot{Y}_{1} + (k_{1} + k_{2}) Y_{1} - k_{2} Y_{2} m_{2} \ddot{Y}_{2} + k_{2} Y_{2} - k_{2} Y_{1} = 0, = 0, = - m_{1} g, = - m_{2} g .

The two spring forces are equal and opposite on the two masses for the connecting spring, as required by Newton's third law.

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