Recent content by Xavicamps

-w^2*sin(Θsin(ωt)+w*t), unless this is the solution I do not get your point

substitution of alpha by Θsin(ωt), is not what we said it was correct in post #17?

yes, the final result would be -w^2*sin(Θsin(ωt)) where Θ would be the elongation of the system

I think in our case alpha should be equal to Θsin(ωt), as this formula is telling us what is the position of the system for every time unit.

Definetely the mistake is mine. we do not have the angular velocity, but the "frequency w". Therefore, I think we arrived to the solution we were looking for: -w^2*sin(w*t+alpha), where we know the angle, and the frequency w. Is that correct?

the angular velocity is Θ*w*cos(w*t) and the angular acceleration is -Θ*w^2*sin(w*t). I think alpha should be substituted by alpha initial + w*t, this is the only solution I can think about

yes, you are right, i made a mistake while saying that w was angular velocity, w is the frequency. Therefore we will obtain that acceleration(angular)=-w^2*sin(alpha), now alpha would be substituted by... angular velocity*time?

Here you have the differential eqution, my problem is if in this case the W is sqrt (g/l+k/m)

Yes, it is always effective

As I can see the result of my "picture" is not the one I was expecting, assume that the rod goes to the ball, and you will see it clear

Is a pendulum with a fixed length rod and a spring connected to the same mass, in horizontal position. I will try to picture it. \ \ \ ><><><><● where: \ rod, ● mass and >< spring

Homework Statement Spring connected to a pendulum holding a ballwith mass m, length l, spring constant k, spring deflection x, angle of pendulum alpha, angular velocity w.Homework Equations Derive the angular acceleration The Attempt at a Solution I made the three body diagram, and find that...