Mass-Spring Oscillator Time-Domain Solution

Consider now the mass-spring oscillator:

Electrical equivalent-circuit:

Newton's second law of motion:

$$f_m(t)=m{\ddot x}(t).$$

Hooke's law for ideal springs:

$$f_k(t)=kx(t)$$

Newton's third law of motion:

$$\begin{eqnarray*} f_m(t) + f_k(t) &=& 0\\ \Rightarrow\; m {\ddot x}(t) + k x(t) &=& 0 \end{eqnarray*}$$

We have thus derived a second-order differential equation governing the motion of the mass and spring. (Note that $x(t)$ is both the position of the mass and compression of the spring at time $t$.)

Taking the Laplace transform of both sides of this differential equation gives

$$\begin{eqnarray*} 0 &=& {\cal L}_s\{m{\ddot x}+ k x\} \\ &=& m{\cal L}_s\{{\ddo... ....~thm again)} \\ &=& ms^2 X(s) - msx(0) - m{\dot x}(0) + k X(s) \end{eqnarray*}$$

Let $x(0)=x_0$ and ${\dot x}(0)={\dot x}_0=v_0$ for simplicity.
Solving for $X(s)$ gives

$$\begin{eqnarray*} X(s) &=& \frac{sx_0 + v_0}{s^2 + \frac{k}{m}} \;\mathrel{\sta... ...m{\Delta}}{=}}\; \tan^{-1}\left(\frac{v_0}{{\omega_0}x_0}\right) \end{eqnarray*}$$

denoting the modulus and angle of the pole residue $r$, respectively.