Undamped Harmonic Motion of a rod

[Feodalherren]

ah yeah, duh!

x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t) - mg/k

 

[Feodalherren]

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

 

[haruspex]

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, that only works at t=0. Over time, the average of that sum of trig functions is zero, not -mg/k.

 

[Feodalherren]

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?

 

[haruspex]

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?

No, you have two. At time zero it is at rest in the equilibrium position.