That's right.. but exactly which trig law do I use and how do I use it to go from equation (1) above in the attached picture to x(t) = Asin(wt + ø).
Or in other words HOW do I go FROM x(t)=x0cos(wt)+(v0/w)sin(wt) ---- (w being = sqrt(k/m) TO x(t) = Asin(wt + ø)
how? HOW? HOW?!??!? HOW?!?
I'm supposed to derive x as a function of time for a simple
harmonic oscillator (ie, a spring). According to my textbook
this is done by using Newton's second law and hooke's law
as this: ma=-kx and one gets a differential equation in
the second order. I can follow the calculations until...
Yeah, you're right.. but one SHOULD be able solve it by puting it up as a second order right? I just want to know what I'm doing wrong, it really bugs me.
Homework Statement
A particle is dropped from rest, at the surface, into a tank containing oil
The acceleration of the particle in the oil is a = g – kv
where g is the gravitational acceleration and –kv being denoted by
the resistance put on the particle by the oil.
Solve for x as a...
Let's say that I solve a inhomogeneous differential equation of
the type d2y/dx2+k*dy/dx = g (k and g being constants)
..and I get the complementary function: y = A + Be^-kx
What would the suggested form of the particular solution be? "Cx" ?