Recent content by Salvador_

x = c*cos(k1/2t) + vo*sin(k.5t)/k1/2 Would I be allowed to set c=0 to make x(0) = 0 so it'll then just be: X = Vo*sin(k.5t)/k1/2

Wait I think this is wrong because a isn't equal to -kx then.

I'm thinking it's: x = cos(kt)/k + v0sin(kt)/k -1/k So when t=0 , x=0 and v=v0

Oh yes I I yes I recognise it's cos and sin. Thanks all.

Oh ok, I've been trying to solve it in first order because I haven't been introduced to second order yet.

I set it up as dv/dx so I could integrate both sides (with respect to position on one side and velocity on the other. v* dv/dx = dx/dt * dv/dx so the dx's cancel out to get dv/dt. No I've been using Morin's CM book and have only been introduced to first order DE's so far.

I set it up as dv/dx so I could integrate both sides (with respect to position on one side and velocity on the other. v* dv/dx = dx/dt * dv/dx so the dx's cancel out to get dv/dt. The book I've been using (Morin) was introducing me to this and I haven't done second order DE's yet.

Sorry but I just don't know how to solve it if the left side is with respect to time and the other side is position.

Then I'd have dv/dt = -kx dv = -kx*dt I'm guessing that's not what you meant because the integral doesn't make sense.

How would I go about doing this? I'm not sure how to solve in time if the function is of position.

Homework Statement A particle of mass m is subject to a force F (x) = -kx. The initial position is zero, and the initial speed is v0. Find x(t). Homework Equations F = m*v*dv/dx = -kx v = dx/dt The Attempt at a Solution I'm new to differential equations, so please excuse me if I make any...

Yes I kind of understand it better now. Thanks!

How are the tensions equal in magnitude when the bead is sliding, shouldn't the upper string have more tension because of the gravity on the bead?

I posted two separate questions in my post. That picture applies only to the second question. The first question is a right angled triangle with a ring traveling in a horizontal circle. The only way I could get the answer 294^(1/2) was assuming that the tensions were the same.

I was able to follow you up to the last two lines. In the original question, the ring is traveling in a horizontal circle. qr is parallel with the ground and pr is at an angle. I'm still unsure as to how the tension is the same and how the ring is "hanging straight down".