1)

*A 5 lb. weight hangs vertically on a spring whose spring constant k = 10. The weight is pulled 6 inches farther down and released. Find the equation of motion, its period and the frequency. What would the units be for k?*
Solution:

We have to keep in mind that British units are being used here so the acceleration of gravity, g = 32 ft/sec

^{2}and the mass m is in slugs. The mass, recall is W/ g so that:

Mass
m = 5 lbs/ 32 ft/sec

^{2 }= 5/32 slug
We’re
given that k = 10 so the initial equation of motion is written:

mx” + 10x = mg

Now,
divide through by the mass m (remember its value):

x” + 10 x/ (5/32 sl) = 32

or: x” + 64 x = 32

Recall
in this harmonic motion eqn. the angular frequency is derived from:

w

^{2}= 64 so that w = Ö64 = 8
Or:
w = 2pf
= 8 and f = 8/ 2p =4/p
rad/sec

The
period is just: T = 2p/w =
2p/ 8 = p/ 4
sec

Getting
back to the solution: Given the above we may write:

x
= c1 sin 8t + c2 cos 8t + ½

(Note
that the 6” has been converted to feet which marks the added extension)

Now,
before the extra 6” pull down what is the natural extension? Well, since the
spring constant k = 10 i.e. 10lbs stretches the spring 1 ft. then 5 lbs. must
stretch it 6” or ½’.

Therefore,
at t = 0 at the instant of pull down we have a total extension of 1 ft so:

1
ft = c1 sin 8(0) + c2 cos (0) + ½

Or:
1 ft = 0 + c2 + ½

So:
c2 = ½

Then
the solution is: x = ½ cos 8t + ½

What
are the units of k? From the preceding
information, k is in lb/ft.

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