8B (Classical Dynamics)

L = \frac{m}{2} (\overset{y}{˙}_{1}^{2} + \overset{y}{˙}_{3}^{2}) + \frac{M}{2} \overset{y}{˙}_{2}^{2} - \frac{k}{2} [(y_{1} - y_{2})^{2} + (y_{2} - y_{3})^{2}] - g (m y_{1} + M y_{2} + m y_{3}) .

T = 2 m + M

Y = \frac{m y _{1} + M y _{2} + m y _{3}}{T} .

All internal forces cancel, so

\ddot{Y} = - g

and

Y (t) = A + \dot{Y} (0) t - \frac{1}{2} g t^{2} .

If the molecule is released from rest, the middle term vanishes.

b

Since

y_{1} = (Q_{s} + Q_{a}) /2

y_{3} = (Q_{s} - Q_{a}) /2

, elimination of the middle coordinate by the center of mass gives

y_{2} = \frac{T Y - m Q _{s}}{M} .

After substitution,

L = \frac{m}{4} (\dot{Q}_{s}^{2} + \dot{Q}_{a}^{2}) + \frac{( T Y ˙ - m Q ˙ _{s} ) ^{2}}{2 M} - \frac{k T ^{2}}{M ^{2}} (Y - \frac{Q _{s}}{2})^{2} - \frac{k}{4} Q_{a}^{2} - T g Y .

The antisymmetric coordinate is decoupled. Its Euler--Lagrange equation is

\frac{m}{2} \ddot{Q}_{a} + \frac{k}{2} Q_{a} = 0, or \ddot{Q}_{a} + \frac{k}{m} Q_{a} = 0.

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