- #36
WannabeNewton
Science Advisor
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- 551
Perhaps I'm misunderstanding you because I don't immediately see the issue at hand. The final orbital velocity of each mass is ##v_{f} = \frac{l_0v_0}{2l_0 + 4x}## hence ##\frac{mv_0^{2}}{4} = \frac{1}{2}k(2x)^{2} + \frac{ml_0^{2}v_0^{2}}{[2l_0(1 + \frac{2x}{l_0})]^{2}} = 2kx^{2} + \frac{mv_0^{2}}{4}(1 + \frac{2x}{l_0})^{-2}##. The dimensionless quantity ##(1 + \frac{2x}{l_0})^{-2}## can be Taylor expanded to 1st order in ##\frac{x_0}{l_0}## to give ##\frac{mv_0^{2}}{4} = 2kx^{2} + \frac{mv_0^{2}}{4}(1 - \frac{4x}{l_0})## from which the result follows. We have retained both the squared velocity in the kinetic energy and the squared displacement in the spring potential while throwing out the terms ##O([\frac{x}{l_0}]^{2})##.