Question

A 2 lb weight stretches a spring 6 inches in equilibrium. An external force F(t)=sin(8t) lb is applied to the weight, which is released from rest 2 inches below equilibrium. Find its displacement for t>0.

Answer 1

m=2/32=1/16
k=4
---------------
\[\frac{ 1 }{ 16 }y''+4y=\sin8t\]
\[y''+64y=16\sin8t\]
\[y _{p}=Acos8t+Bsin8t\]
\[y'_{p}=-8Asin8t+8Bcos8t\]
\[y''_{p}=-64Acos8t-64Bsin8t\]
Substitute:
\[y''_{p}+64y _{p}=0=16\sin8t\]
Now what?

Answer 2

\(\color{000000#}{\displaystyle \frac{1}{ 16}y''+4y=8\sin t \quad \Longrightarrow \quad y''+64y=128\sin t }\)

The homogeneous solution.
\(\color{000000#}{\displaystyle r^2+64=0 }\)
\(\color{000000#}{\displaystyle r=\pm8i }\)
\(\color{000000#}{\displaystyle y_t=c_1\cos (8t)+c_2\sin (8t) }\)

The general solution.
\(\color{000000#}{\displaystyle y_p=A\sin t }\)
(I know what I'm doing, there is no first derivative in the equation)

\(\color{000000#}{\displaystyle y''+64y=128\sin t }\)
\(\color{000000#}{\displaystyle -A\sin t+64A\sin t=128\sin t }\)
\(\color{000000#}{\displaystyle 63A\sin t=128\sin t }\)
\(\color{000000#}{\displaystyle A=128/63 }\)

The general solution:
\(\color{000000#}{\displaystyle y_t=c_1\cos (8t)+c_2\sin (8t) +(128/63)\sin t }\)

Answer 3

Then, you will have to look more carefully at the problem, and find the initial values to fix \(c_1\) and \(c_2\).

Answer 4

How did you get 8sin(t)?

Answer 5

oh I misread that, sorry:)

Answer 6

that way it's much more work. I don't think I can type all of that right now. I will tell you tho' that the \(y_p\) that you guessed is wrong.

Answer 7

Always find the homogeneous solution first, and then the particular, because if you don't do that, you can get a wrong particular solution, and you will end up just recreating the homogeneous solution (and that is obviously purposeless).

Answer 8

So what should yp be in this case?

Answer 9

Please, find the homogeneous solution first. What is it?

Answer 10

\[y _{h}=C _{1}\cos8t+C _{2}\sin8t\]

Answer 11

Yes, good.

Answer 12

Now, the particular solution?
(Your best guess)

Answer 13

Since we already have sin8t in the homogeneous solution. The particular solution would be yp=Acos8t?

Answer 14

really totally at sea with the [ancient!!] use of imperial measurements, lol!! can't believe the Americans put a man on the moon using this crap.

more seriously, i would need to sit down and make sure the constants that switch between the two flow through in completely linear fashion.

can do that in a while, if that helps. in short term, sorry :-(