1__)__ A 5 lb. weight hangs on a spring whose spring constant is 10. The weight is pulled 6 inches further down and released. Find the equation of motion, its period and frequency. If g = 32 ft/sec/sec give the units for the spring constant.

__Soln.__

**lbs./ ft.**

The form of the equation of motion will then be:

m(d^{2}x/dt ^{2}) + k x = mg

Or: m(d^{2}x/dt ^{2}) + 10 x = mg

And since W= mg = 5 lb. and g = 32 ft/sec/sec

m = 5 lb. / 32 ft/sec/sec

Þ (d^{2}x/dt ^{2}) + (k/ m) x = g

Þ (d^{2}x/dt ^{2}) + 64 x = 32

But note angular frequency:

w = Ö(k/ m) = w = Ö(10 lb./ft/ 5/32 lb) = 8 s^{-1}

So we obtain solution:

x = 1/2 cos w t + 1/2

Or: x = 1/2 cos 8 t + 1/2

(N.B. At t = 0, x = 6 in = 1/2 ft, then pull down another 6 in. or 1/2 ft)

Frequency: f = w / 2 p = 8 s^{-1} / 2 p = 4/ p s^{-1 }

Period = T = 2 p/ w = 2 p/8 s^{-1} = p/4 s

_{o}sin w t

__Soln.__

^{at }= ò e

^{at }E dt + C

_{o}sin w t

_{o}[RL sin w t - w L

^{2}cos w t / (R

^{2}+ w

^{2}L

^{2})] + C e

^{-at }

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