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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$ .

Figure D.4: Equivalent circuit for the mass-spring oscillator.

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:

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)$ 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

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

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 (s) &=& sx_0 + v_0s^2 + km &Delta#Delta;= rs+j&omega#omega;_0 + rs-j&omega#omega;_0,     &omega#omega;_0&Delta#Delta;=k/m,    and

r&=& x_02 + j v_02&omega#omega;_0 &Delta#Delta;= R_r e^j&thetas#theta;_r,    with

R_r &&Delta#Delta;=& v^2_0 + &omega#omega;_0^2 x^2_02&omega#omega;_0,          &thetas#theta;_r &Delta#Delta;= ^-1(v_0&omega#omega;_0x_0)

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

x(t) &=& re^{-j{\omega_0}t} + \overline{r}e^{j{\omega_0}t}
= 2\mbox{re}\left\{re^{-j{\omega_0}t}\right\}
= 2R_r\cos({\omega_0}t - \theta_r)\\
&=& \frac{\sqrt{v^2_0 + {\omega_0}^2 x^2_0}}{{\omega_0}}
\cos\left[{\omega_0}t - \tan^{-1}\left(\frac{v_0}{{\omega_0}x_0}\right)\right].

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

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

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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2016-02-11 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University