Recent content by seanryan

Oh, that's how they did it. Significantly less work than what I was doing. Thanks for all your help.

Ok, so the sign error just makes the potential energy U=k[-2L*(\sqrt{R^{2}+x^{2}+2Rxcos(\alpha)}+\sqrt{R^{2}+ x^{2}-2Rxcos(\alpha)})+2x^{2}] If I expand this I would get -4kLR-2k\frac{L}{R}sin^{2}(\alpha)x^{2} I can account for the 1-L/R by not expanding the x^2 so I have...

I also tried taking the x component out of the force then squaring it and putting that into 1/2kx^2 but that didn't work either

What I first did was imagine that the mass was displaced a distance d to the right. Then I found what the new length of the springs are. For the left 2 springs i got r_{1}^{2}= R^{2}+d^{2}+2Rdcos(\alpha) for the right r_{2}^{2}=R^{2}+d^{2}-2Rdcos(\alpha) I then broke these into vector...

Ok, so the system is a ring with four identical springs connected to a single mass. Each spring is in a different quadrant and at an angle alpha from the horizontal. L is the relaxed length of each spring, k' is the spring constant, and R is the radius of the ring. and ignore gravitational and...