Spring/Mass System: Driven Motion problem

[hank]

(5.1.35)

A mass m is attached to the end of a spring whose constant is k. After the mass reaches equilibrium, its support begins to oscillate vertically about a horizontal line L according to formula h(t). The value of h represents the distance in feet from L.
(a) Determine the Differential Equation of Motion if the entire system moves through a medium offering a dampening force numerically equal to Bx'.
(b) Solve the differential equation in (a) if the spring is stretched 4' by a mass weighing 16 lbs and B = 2, h(2) = 5cost, x(0) = x'(0) = 0.


For a, I get: mx" = -k(x-h) - Bx' => mx" + Bx' + kx = kh

B is where I get stuck. Specifically, I don't know what to do with h(2) = 5cost.

Here's what I get to start:
m = W/g => m = 16/32 => m = 1/2
B = 2
Using Hooke's law, F = ks, I plug in 32 = k2 and get k = 32.

So, I want to set it up like
(1/2)x" + 2x' +16x = 80cost => x" + 4x' + 64x = 160cost

I really don't think this is right.

Specifically, I think I'm treating the h(2) = 5cost incorrectly.
In previous problems, it would have read h(0) = 5cost.
What do I do with h(2)?

 

[Unco]

If h is a function of t, then h(2) should be a number; e.g., if h(t) = t^2, then h(2) = 4. So h(2) = 5cos(t) doesn't make sense; are you sure you have this written correctly? My guess is it would just be h(t)=5cos(t). And one minor point: 16*2=32, not 64.

 

[hank]

Well, the book definitely says h(2).
It could very well be a typo.

Also, thanks for the catch on my multiplication.
I'll try the problem again with those things in mind.

Thanks!