Recent content by splatcat

No that is not what I did, I assumed you were measuring a tiny change in v and x instead of x and t.

They are, I was trying to show someone that they could do what I asked in the initial question with a different derivative, they were confused and I was struggling to justify that it was true, getting myself in a loop of confusion. I was not really asking people to do it for me.

surely you just get = 1/t then intergrate both sides you get v = x/t which is correct, so dx/dv must be = to t ?? (thank you for your reply, :) )

Does dx/dv = t ? Can you just manipulate equations like this? [ x is position and v is velocity, t time :P ] Stu

Thank you very much for your help. I am pretty sure I understand where I have gone wrong :)

(Voltage amplitudes was the quesion :) ) Sorry I did mean to put a minus sign in there instead of a + sign. However does the system not work like a potential divider? Only resistances are impedances? Stu

thank you for your response, your question helped direct my reading. I think I have a better idea of what I am doing now. I have come to a new solution. Is it any closer? I have some understanding of complex impedance now. w = omega Vout/Vin = (R + iLw) / (R + i(wL + 1/wC) )

thank you for any help in advance. The question is; In a series LRC circuit, find an equation that shows the ratio of Voltage out [the voltage measured over the capacitor] over the voltage in from an A/C source as a function of L R C and omega Starting with this equation: Vs = sqrt(VR2...