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First, let's review the basic ideas of simple harmonic motion (I'm assuming an early university level). Starting with Newton's equation: $$F=ma$$ and using Hooke's law $$ma=-kx$$ then recognizing that acceleration is the second derivative of position x $$mx''= -kx$$ We know that simple harmonic motion is sinusoidal, so we substitute $x=\sin(\omega t)$ ...

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Note: I see there's been quite a bit of activity while I was typing this up. I'll leave the answer here for now. it doesn't seem immediately intuitive how the problem works Assume the spring provides a linear, tangential restoring force for small angular displacement (rotation) of the disk. The intuition here is that the problem will 'look' like an ...

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Once the mass is released, the center of mass will move at a constant velocity. Superposed on that is the relative motion of the two masses - first towards each other, then away. They will be in exact antiphase so the center of mass has constant velocity. Your mistake was to set x up as a cosine function - that implies that it is at an extreme of position ...

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I don't know if you have already done integration in high school but at least here is a solution without using multivariable calc. The kinetic energy of the spring with length $L$ is the following $$E_{kin}=\int_m \frac{u^2}{2} \, \text{d}m$$ where $u$ is the speed of a little mass element d$m$ and the integration is over the mass of the spring. Assuming ...

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Doesn't the result follow from a simple diagram? If you change the X,Y position by a small amount dx, dy, the change in length of the hypotenuse is simply \begin{align} &= \sqrt{(a+dx)^2 + (a+dy)^2} - \sqrt{a^2 + a^2}\\ &= \sqrt{2a^2 + 2a dx + 2a dy + ...} - \sqrt{2a^2}\\ &= a\sqrt{2}\left(1 + \frac{dx}{a} + \frac{dy}{a} - 1\right)\\ &= ...

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Variations of this problem show up all the time. If you start with the spring "locked" and the spheres charged, then release the spring, it will expand to the new length and when it gets there the spheres will have a velocity - essentially you have a simple harmonic oscillator and the point of (new) equilibrium is the point where the oscillator moved ...

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