# Homework Help: Solving a Differential Equation from story problem

1. May 15, 2010

### TG3

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
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?

2. May 15, 2010

### Staff: Mentor

There are also two initial conditions. You should state them with your differential equation.
Well, really it does, since it makes the difference between complex solutions for r and real solutions.

In any case,
Be consistent in your variables. The dependent variable started off as u; now it's y.

It all depends on which direction you choose to be the positive y axis. I would be more inclined to say that y(0) = -3.
It could be that the book's answer has rationalized the radical. What answer does the book give?

3. May 15, 2010