Damped harmonic oscillator, no clue

[Smity]

Homework Statement

I have a ball of 20 kg describing a damped harmonic movement, ie,
m*∂^2(x)+R*∂x+K*x=0,
with m=mass, R=resistance, K=spring constant.
The initial position is x(0)=1, the initial velocity is v(0)=0.
Knowing that v(1)=0.5, v(2)=0.3, I have to calculate K and R.

2. The attempt at a solution

I know that if R^2 < 4*m*K, the solution with x(0)=1 and v(0)=0 is such that:
∂x(t)=exp(-R/(2*m)*t)*[-(R/(2*m)^2)/(√[K/m-(R/(2*m))^2])-√[K/m-(R/(2*m))^2]]*sin(√[K/m-(R/(2*m))^2]*t), and I solve the sistem of equations, but it has to be a simpler way to do it (and also I don't use the mass of the ball)

Thanks!

 

[rude man]

First, realize that you may have an underdamped or overdamped system. The criterion for underdamped is ζ < 1. For this case, rewrite your equation as follows: (m/K)x" + (R/K)x' + x = 0

Let
m/K = 1/ω₁²
R/K = 2ζ/ω₁
so your equation becomes, in "standard" form,
(1/ω₁²) x'' + 2ζ/ω₁ x' + x = 0
with initial condition x(0+) = 1
with solution

x(t) = ω₁²exp[(-ζω₁t/√(1-ζ²)] sin[ω₁√(1-ζ²)t + ψ],
where ψ = arc tan √(1-ζ²)/(-ζ).

This is the solution of the "standardized" equation above.
Now take (d/dt) of this to get x'(t),
then impose x'(1) = 0.5 and x'(2) = 0.3.

If you succed in getting positive real numbers for R and K you are done.

But it's possible you may have an overdamped (or critically damped) case, in which case rewrite your equation as

T₁T₂ x'' + (T₁ + T₂) x' + x = 0
for which the solution would be

1/T₁T₂(T₁ - T₂) [T₁exp(-t/T₂) - T₂exp(-t/T1)].

Again, take x'(t) from that and force the two conditions on x'(1) and x'(2).

You have some messy math ahead of you and I don't see a simple way to avoid it unless you can find a pre-cooked solution somewhere. I don't know any such place.
.