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Assume that there's no damping, an object stretches a spring 4 inches in equilibrium. Find its displacement for t>0 if it's initially displaced 36 inches above equilibrium and given a downward velocity of 2 ft/s.
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\[y''+\frac{ k }{ m }y=0\]
That above is the equation of motion but how do I find k and m?
@Kainui @perl @jim_thompson5910
the spring stiffness and object inertia simply determine the frequency of the response: $$y''+\omega^2y=0;\quad\omega^2=\frac{k}m$$which has the general solution $$y(t)=A\cos(\omega t+\phi)$$
we're told \(y(0)=3, y'(0)=-2\) since 36 in is 3 ft which gives $$A\cos(\phi)=3\\-A\omega \sin(\phi)=-2$$ so $$A^2=(A\cos(\phi))^2+(-A\sin(\phi))^2=9+4/\omega^2\\\tan\phi=\frac{A\sin\phi}{A\cos\phi}=\frac2{3\omega}$$
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