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Mass-Spring Oscillator Time-Domain Solution

Consider now the mass-spring oscillator:


Electrical equivalent-circuit:


Newton's second law of motion:

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

Hooke's law for ideal springs:

$\displaystyle f_k(t)=kx(t)

Newton's third law of motion:

f_m(t) + f_k(t) &=& 0\\
\Rightarrow\; m {\ddot x}(t) + k x(t) &=& 0

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

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)

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

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
\\ [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)

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

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``The Laplace Transform'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2015-03-30 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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