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

$$\begin{eqnarray*} 0 &=& {\cal L}_s\{m{\ddot x}+ k x\} \\ &=& m{\cal L}_s\{{\ddot x}\} + k {\cal L}_s\{x\} \quad \hbox{(linearity)} \\ &=& m\left[s{\cal L}_s\{{\dot x}\} - {\dot x}(0)\right] + k X(s) \quad\hbox{(differentiation theorem)} \\ &=& m\left\{s\left[sX(s) - x(0)\right] - {\dot x}(0)\right\} + k X(s) \quad \hbox{(diff.~thm again)} \\ &=& ms^2 X(s) - msx(0) - m{\dot x}(0) + k X(s) \end{eqnarray*}$$

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

$$\begin{eqnarray*} X(s) &=& \frac{sx_0 + v_0}{s^2 + \frac{k}{m}} \;\mathrel{\stackrel{\mathrm{\Delta}}{=}}\; \frac{r}{s+j{\omega_0}} + \frac{\overline{r}}{s-j{\omega_0}},\quad {\omega_0}\mathrel{\stackrel{\mathrm{\Delta}}{=}}\sqrt{k/m}, \\ [10pt] r&=& \frac{x_0}{2} + j \frac{v_0}{2{\omega_0}} \;\mathrel{\stackrel{\mathrm{\Delta}}{=}}\; R_r e^{j\theta_r},\quad\hbox{with}\\ [10pt] R_r &\mathrel{\stackrel{\mathrm{\Delta}}{=}}& \frac{\sqrt{v^2_0 + {\omega_0}^2 x^2_0}}{2{\omega_0}}, \qquad \theta_r \;\mathrel{\stackrel{\mathrm{\Delta}}{=}}\; \tan^{-1}\left(\frac{v_0}{{\omega_0}x_0}\right) \end{eqnarray*}$$

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