Bryon0101 wrote:
Is the approach correct for this sort of problem?
A 32 lb object stretches a spring 8 ft. If the resistive force due to damping is 5x' , determine the displacement function if the object is released from 1 ft below the equilibrium position with an upward velocity of 1 ft/sec.
Here is what I have:
w = m(32ft/s)
f = ma
f = k(s + x)
mx'' + cx' + kx = 0
x(0) = 8ft
x'(0) = 1ft/s
c = 5
w = m(32ft/s): 32lb = m(32ft/s) m = 1 slug
f = k(x + x): f = kx 32 = k*8 k = 4
1x'' + 5x' + 4x = 0
So I took the Laplace transform of that and got this:
s^(2)X - sx(0) - x'(0) + 5[sX - x(0)] + 4X = 0
s^(2)X - 8s - 1 + 5[sX - 1] + 4X = 0
s^(2)X + 5sX +4X - 8x - 2 = 0
X(s^2 + 5s + 4) = 8s + 2
X = (8s + 2)/(s+4)(s+1)
The inverse Laplace of X yeilded 10e^(-4t) - 2e^(-t)
Thanks!
No. Note the initial condition highlighted in red, which your answer fails to satisfy.