Damped spring problem with laplace transform

1. Nov 9, 2014

Sneakatone

1. The problem statement, all variables and given/known data
A 4-pound weight stretches a spring 2 feet. The weight is released from rest 15 inches above the equilibrium position, and the resulting motion takes place in a medium offering a damping force numerically equal to 7/8 times the instantaneous velocity. Use the Laplace transform to find the equation of motion x(t). (Use g = 32 ft/s2 for the acceleration due to gravity.)

2. Relevant equations
my''+(beta)y'+ky=0

m=4/32=1/8
beta=7/8
k=4/2=2
3. The attempt at a solution
1/8y''+7/8y'+2y=0, y(0)=-18, y'(0)=0

I know the answer which is:
-1/10e^(-7t/2)[7√(15)sin(√(15)t/2)+15cos(√(15)t/2)]
https://www.webassign.net/latexImages/5/f/f1c0fb013b3d1d7c4fddd1209a1d60.gif

but my awnser came out to be:
http://www4c.wolframalpha.com/Calculate/MSP/MSP104701e9a70abfi51f43a000045edi0bi5e950fi2?MSPStoreType=image/gif&s=14&w=373.&h=45 [Broken].

which makes me fell like my y'(0) is incorrect. I would think that y'(0) would be a valid initial condition but the problem seem to not state anymore about velocity.

Last edited by a moderator: May 7, 2017
2. Nov 9, 2014

vela

Staff Emeritus
I take it the weight is released from 18 inches above the equilibrium point, not 15 inches like you typed above. The link to your answer doesn't work, but I suspect the problem is simply that you didn't specify y(0) in the correct units. The statement does say that the weight was released from rest, so your condition of y'(0) = 0 is correct.

Last edited by a moderator: May 7, 2017
3. Nov 9, 2014

Sneakatone

my answer came out to be

y(x) = -6/5 e^(-7 x/2) (7 sqrt(15) sin((sqrt(15) x)/2)+15 cos((sqrt(15) x)/2))
which is different and yes the problem is suppose to be released at 18 not 15.

4. Nov 9, 2014

Sneakatone

my answer came out to be
y(x) = -6/5 e^(-7 x/2) (7 sqrt(15) sin((sqrt(15) x)/2)+15 cos((sqrt(15) x)/2))
which is different. and yes the problem is suppose to be released at 18 not 15.

5. Nov 9, 2014

vela

Staff Emeritus
Unless you show your work, we can't figure out what you're doing wrong.

6. Dec 7, 2014

Sneakatone

sorry for the late reply but I found that the released spring distance was in inches so I just had to convert the 15 in to ft.