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Mass-Spring Oscillator Analysis

Consider now the mass-spring oscillator depicted physically in Fig.D.3, and in equivalent-circuit form in Fig.D.4.

Figure D.3: An ideal mass $ m$ sliding on a frictionless surface, attached via an ideal spring $ k$ to a rigid wall. The spring is at rest when the mass is centered at $ x=0$.
\includegraphics{eps/massspringwall}

Figure D.4: Equivalent circuit for the mass-spring oscillator.
\begin{figure}\input fig/tankec.pstex_t
\end{figure}

By Newton's second law of motion, the force $ f_m(t)$ applied to a mass equals its mass times its acceleration:

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

By Hooke's law for ideal springs, the compression force $ f_k(t)$ applied to a spring is equal to the spring constant $ k$ times the displacement $ x(t)$:

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

By Newton's third law of motion (``every action produces an equal and opposite reaction''), we have $ f_k = -f_m$. That is, the compression force $ f_k$ applied by the mass to the spring is equal and opposite to the accelerating force $ f_m$ exerted in the negative-$ x$ direction by the spring on the mass. In other words, the forces at the mass-spring contact-point sum to zero:

\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)$ in Fig.D.3 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...
...orem again)} \\
&=& ms^2 X(s) - msx(0) - m{\dot x}(0) + k X(s).
\end{eqnarray*}

To simplify notation, denote the initial position and velocity by $ x(0)=x_0$ and $ {\dot x}(0)={\dot x}_0=v_0$, respectively. Solving for $ X(s)$ gives

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

denoting the modulus and angle of the pole residue $ r$, respectively. From §D.1, the inverse Laplace transform of $ 1/(s+a)$ is $ e^{-at}u(t)$, where $ u(t)$ is the Heaviside unit step function at time 0. Then by linearity, the solution for the motion of the mass is

\begin{eqnarray*}
x(t) &=& re^{-j{\omega_0}t} + \overline{r}e^{j{\omega_0}t}
= ...
...ga_0}t - \tan^{-1}\left(\frac{v_0}{{\omega_0}x_0}\right)\right].
\end{eqnarray*}

If the initial velocity is zero ($ v_0=0$), the above formula reduces to $ x(t) = x_0\cos({\omega_0}t)$ and the mass simply oscillates sinusoidally at frequency $ {\omega_0}=
\sqrt{k/m}$, starting from its initial position $ x_0$. If instead the initial position is $ x_0=0$, we obtain

\begin{eqnarray*}
x(t) &=& \frac{v_0}{{\omega_0}}\sin({\omega_0}t)\\
\;\Rightarrow\; v(t) &=& v_0\cos({\omega_0}t).
\end{eqnarray*}


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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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