- #1
TG3
- 66
- 0
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?
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?