Feodalherren
- 604
- 6
ah yeah, duh!
x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t) - mg/k
x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t) - mg/k
No, that only works at t=0. Over time, the average of that sum of trig functions is zero, not -mg/k.Feodalherren said:Or acutally, couldn't I just find that with
x(0)=-mg/k= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t)
so
A=-mg/k
No, you have two. At time zero it is at rest in the equilibrium position.Feodalherren said:So then by the principle of superpositioning it should be
x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t) - mg/k
but then I arrive back at my original issue - how do I find A and B. I only have one initial condition?