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...