Recent content by ethan123

Thanks for the reply So letting x_0 be the distance from equilibrium. F=kx ma=kx a=\frac{k}{m}x And since \omega =\sqrt{\frac{k}{m}} a=\omega^2x And the x found in part i) would be the distance from the natural length of the spring to equilibrium. Does this make sense?

Homework Statement A block of mass M can slide freely (without friction) on a flat surface held at an angle \theta to the horizontal. The block is attached to a spring of natural length l and spring constant k. The other end of the spring is fixed to a nail driven into the sloping surface...