(adsbygoogle = window.adsbygoogle || []).push({}); The problem statement, all variables and given/known data

A mass weighing 2 lbs stretches a spring 6 inches. If the mass is then pulled down an additional 3 inches and then released, and there is no damping, determine the position u of the mass at ay time T according to the form Acos(wT-phi).

The attempt at a solution

The k of the spring is 2 lbs/ 6 inches, or 1/3 lbs/inches. The mass is 2lbs/32ft/sec^2, or 1/16.

From this I get the differential equation:

1/16U'' +1/3U = 0

Making U e^rt and skipping a few steps

1/16r^2+1/3 = 0

Using the quadratic:

r = 0+- 8 sqroot(1/12) (-1/12 actually, but it doesn't matter here.)

So: y = C1 cosine (8 sqroot(1/12)T) + C2 sine (8 sqroot(1/12)T)

The initial condition of y(0) ought to be 3, since the spring was stretched 3 inches, and y'(0) ought to be 0 since it was released not pushed, correct? (I'm suspicious of this step.)

If so, y(0) = C1 cosine (8 sqroot(1/12)T) + 0 (since the sine of 0 is 0).

And furthermore, y(0) = C1, so C1 =3.

y'(0) = [0] + C2 (8 sqroot(1/12)) cosine (8 sqroot(1/12)T)

y'(0) = C2 (8 sqroot (1/12))

0 = C2

So y=3cosine(8sqroot(1/12)T)

From this, translating to the form of A cosine (wT-phi) isn't hard:

A = (3^2)^.5 = 3

W = 8 sqroot(1/12)

And phi is tan^-1 (0/3) = 0

My solution is 3 cosine 8(sqroot(1/12))T.

Unfortunately, this is wrong. Where did I goof up?

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# Homework Help: Solving a Differential Equation from story problem

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