Modeling mass-spring-dashpot system

[Kavorka]

Homework Statement

Suppose that the mass in a mass-spring-dashpot system with m = 25, c = 10, and k = 226 is set in motion with x(0) = 2 and x'(0) = 4. Find the smallest time t>0 for which x(t) = 0

Homework Equations

mx'' + cx' + kx = 0

The Attempt at a Solution

I rushed through this a bit due to lack of time today so there may be mistakes, I just need to be sure I'm doing this correctly.

The differential equation is: 25x'' + 10x' + 226x = 0
Which has the characteristic equation: 25r² + 10r + 226 = 0
with roots: r = -(1/5) +/- 3i

Giving the general solution:
x(t) = e^-(1/5)t(Acos(3t) + Bsin(3t))

Plugging in the initial conditions:
x(0) = 2 = A

x'(t) = -(1/5)Ae^-(1/5)tcos(3t) -3A e^-(1/5)tsin(3t) - (1/5)e^-(1/5)tsin(3t) + 3Be^-(1/5)tcos(3t)

x'(0) = 4 = -(1/5)(2) + 3B
B = 22/15

Finding combined solution:
x(t) = Ce^-(1/5)tcos(3t-a)

C = (A² + B²)^1/2 = 3(346)^1/2/15

a = tan^-1(B/A) = 0.6327 radians

x(t) = (3(346)^1/2/15)e^-(1/5)tcos(3t - 0.6327)

If this is right (it seems too messy, might have messed up), how would I solve it when equal to zero?

 

[BvU]

Giving the general solution:
x(t) = e^-(1/5)t(Acos(3t) + Bsin(3t))

For which you find A and B correctly A correctly.The first factor is > 0, so basically all you want to solve is

Acos(3t) + Bsin(3t) = 0​

And that's not so hard.

And if you want the other form, same thing: solving cos(3t - atan(B/A)) = 0 should be enough -- and give the same result

Something's wrong with your general solution, though, so fix that first !

 

[Kavorka]

I checked again, and with Wolfram, and I did get A and B right. C should be 2(346)^0.5 /15 though. When I solve cos(3t - 0.6327) = 0, I get t = 0.7345 which isn't right. This answer x2 is right though, and I'm not sure why.

Edit: misunderstanding, my answer is right.

 

[BvU]

Yeah, B was 62/45. Well done!