# Can you solve this problem in Mechanics?

Nov 16, 2016

$x \left(t\right) = \left({x}_{e} - \frac{g m}{k} \tan \left(a\right)\right) \left(\cos \left(\sqrt{\frac{k}{m}} t\right) - 1\right)$

#### Explanation:

Considering the movement projected over the ramp, applying Newton's second law and making $\alpha = \frac{\mathrm{dy}}{\mathrm{dx}} = a$

$- m g \sin \left(\alpha\right) - k \left(x - {x}_{e}\right) \cos \left(\alpha\right) = m \ddot{x} \cos \left(\alpha\right)$ or

$\ddot{x} + \frac{k}{m} x + g \tan \left(\alpha\right) - k {x}_{e} \cos \left(\alpha\right) = 0$

This second order linear differential equation has the general solution.

x(t)=x_e+C_1 Cos(sqrt[k/m] t) + C_2 sin(sqrt[k/m] t) - ( g m tan(alpha))/k

${C}_{1} , {C}_{2}$ are determined according to the initial conditions:

$\left\{\begin{matrix}x \left(0\right) = {x}_{0} \\ \dot{x} \left(0\right) = 0\end{matrix}\right.$

so finally

$x \left(t\right) = \left({x}_{e} - \frac{g m}{k} \tan \left(a\right)\right) \left(\cos \left(\sqrt{\frac{k}{m}} t\right) - 1\right)$

The $y \left(t\right)$ position formulation, is let as an exercise to the reader.