Damped harmonic oscillator, no clue!

1. Feb 16, 2013

Smity

1. The problem statement, all variables and given/known data

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!

2. Feb 16, 2013

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.
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