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Mechanical Impedance Analysis

Impedance analysis is commonly used to analyze electrical circuits [110]. By means of equivalent circuits, we can use the same analysis methods for mechanical systems.


For example, referring to Fig.7.9, the Laplace transform of the force on the spring $ k$ is given by the so-called voltage divider relation:8.2

$\displaystyle F_k(s)
= F_{\mbox{ext}}(s) \frac{R_k(s)}{R_m(s)+R_k(s)}
= F_{\mbox{ext}}(s) \frac{\frac{k}{s}}{ms+\frac{k}{s}}
$

Similarly, the Laplace transform of the force on the mass $ m$ is given by

$\displaystyle F_m(s) = F_{\mbox{ext}}(s) \frac{R_m(s)}{R_m(s)+R_k(s)} = F_{\mbox{ext}}(s) \frac{ms}{ms+\frac{k}{s}}. \protect$ (8.1)

As a simple application, let's find the motion of the mass $ m$, after time zero, given that the input force is an impulse at time 0:

$\displaystyle f_{\mbox{ext}}(t)=\delta(t) \;\leftrightarrow\; F_{\mbox{ext}}(s)=1
$

Then, by the ``voltage divider'' relation Eq.$ \,$(7.1), the Laplace transform of the mass force $ f_m(t)$ after time 0 is given by

$\displaystyle F_m(s) = \frac{ms}{ms+\frac{k}{s}}
= \frac{s^2}{s^2+\frac{k}{m}}
\isdef \frac{s^2}{s^2+\omega_0^2},
$

where we have defined $ \omega_0^2\isdef k/m$. The mass velocity Laplace transform is then
\begin{eqnarray*}
V_m(s) &=& \frac{F_m(s)}{ms} \;=\; \frac{1}{m} \cdot \frac{s}{...
...}\right]\\ [5pt]
&\leftrightarrow& \frac{1}{m} \cos(\omega_0 t).
\end{eqnarray*}
Thus, the impulse response of the mass oscillates sinusoidally with radian frequency $ \omega_0=\sqrt{k/m}$, and amplitude $ 1/m$. The velocity starts out maximum at time $ t=0$, which makes physical sense. Also, the momentum transferred to the mass at time 0 is $ m\,v(0+) = 1$; this is also expected physically because the time-integral of the applied force is 1 (the area under any impulse $ \delta(t)$ is 1).
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