Solving a Differential Equation from story problem

[TG3]

Homework Statement

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?

 

[Mark44]

Homework Statement

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

There are also two initial conditions. You should state them with your differential equation.

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.)

Well, really it does, since it makes the difference between complex solutions for r and real solutions.

In any case,

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

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

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.)

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.

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?

It could be that the book's answer has rationalized the radical. What answer does the book give?

 

[TG3]

The answer given is:
.25 cos(8t)

 

[Mark44]

Your spring constant is incorrect, which leads to an incorrect differential equation. The mass is 1/16 slug (1 slug = 32 lb/(ft/sec)^2). Your spring constant is 2lb/6 in = 4 lb/ft. The spring constant should not be in terms of inches.

All length measures, including the initial condition, should be in terms of feet, not inches.