Originally Posted by

**billym** A block of mass $\displaystyle m$ lies on a smooth plane $\displaystyle AB$ inclined at an angle $\displaystyle \alpha$ to the horizontal. The block is attached to $\displaystyle A$ and $\displaystyle B$ by a spring on either side (up the slope is positive).

The general solution for the equation of motion is:

$\displaystyle

x\left ( t \right ) = A~cos \left ( t \sqrt{\frac{5k}{m}} + \phi \right )

$

Given:

$\displaystyle

x\left ( 0 \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha)

$

I have the particular solution as:

$\displaystyle

x\left ( t \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) cos \left ( t \sqrt{\frac{5k}{m}} \right )

$