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Amplitude of spring after collision
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[QUOTE="andrewkirk, post: 5472316, member: 265790"] Your calculations look correct to me. Perhaps the book has a mistake. They sometimes do. The formula ##v=\frac{2\pi}T\times A## gives the velocity v of an object in simple harmonic motion (SHM) when it passes the equilibrium point. It is easily derived when you consider that the formula for SHM is $$x(t)=A\sin(2\pi t/T)$$ SHM is what will be happening after the collision, and the point of the collision is the equilibrium point, as that is where the spring is under no strain in either direction, and velocity is maximum. Even though I don't understand much Indonesian/Malaysian, I can see what the book is doing. They use the formula ##T=2\pi\sqrt{\frac{mass}k}## for the period of an object in SHM, together with the other formula, to solve for ##A##. They correctly set ##mass=2m##. However they make a mistake with ##v##. They should use ##v'=v/2## instead, because that's what the velocity is after the collision. So you are correct and the book made a mistake. bagus! [/QUOTE]
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Amplitude of spring after collision
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