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A 1 lb weight is suspended from a spring. Let y give the deflection (in inches) of the weight from its static deflection position, where “up” is the positive direction for y. If the static deflection is 24 in, find a differential equation for y. Solve, and determine the period and frequency of the SHM of the weight if it is set in motion.

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Well, $F=kx$ for spring. Since static equilbrium balances against gravity $kx=mg$. On the other hand, when in motion the constant force of gravity does not effect the oscillation frequency and we can write $m\ddot{y} +ky=0$ which gives characteristic equation $m\lambda^2+k=0$ hence $\lambda = \pm i \sqrt{k/m}$. Let $\omega = \sqrt{k/m}$ and find $y(t) = A\cos(\omega t + \phi)$ as the solution to the equations of motion. The period $T$ is defined such that $\omega T = 2\pi$ hence $T = 2\pi /\omega$. You are given $k(24)=1$ in inches and lbs. To finish you need to deal with the difference between $m$ and $mg$. Beware the inches vs. ft issue. Good luck. (if you believe in that sort of thing)

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So the differential equation I need to solve is simply $my'' + ky = 0; y(24) = 1$? –  user1038665 Oct 25 '12 at 3:24
    
I'm curious as to how the condition $k(24) = 1$ allows me to to come up with a solution to the differential equation. –  user1038665 Oct 25 '12 at 6:55
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