Engineering at Alberta Courses » Response of Spring–Mass System to an Exponential Decay

Transient Vibrations: Response of Spring–Mass System to an Exponential Decay

Exponential Decay

The equation of motion in this case is

$\begin{equation*} m \ddot{x} + k x = F_0 e^{-a t}, \end{equation*}$

for which the particular solution can be shown to be

$\begin{equation*} x_P(t) = \frac{F_0}{m a^2 + k} e^{-a t} \end{equation*}$

so that the total solution becomes

(7.7) $\begin{equation*} \label{eq:ExponentialDecayResponseGeneral} x(t) = A \sin{pt} + B \cos{pt} + \frac{F_0}{m a^2 + k} e^{-a t}. \end{equation*}$

If the initial conditions are

$\begin{equation*} x(0) = 0, \qquad \dot{x}(0) = 0, \end{equation*}$

then

$\begin{equation*} A= \frac{F_0 a}{\bigl(m a^2 + k\bigr) p}, \qquad B= \frac{-F_0}{m a^2 + k}, \end{equation*}$

and we get

(7.8) $\begin{equation*} \label{eq:ExponentialDecayResponseIC} x(t) = \frac{F_0}{m a^2 + k} \biggl[ \frac{a}{p} \sin{pt} - \cos{pt} + e^{-at} \biggr]. \end{equation*}$

$\begin{equation*} \boxed{x(t) = \frac{F_0}{k} \frac{1}{1 + \bigl( \frac{a}{p}\bigr)^2} \biggl[\frac{a}{p} \sin{pt} - \cos{pt} + e^{-at} \biggr].} \end{equation*}$

Response to Exponential Decay Function with $x(0)$ = $\dot{x}(0)$ = 0

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Transient Vibrations: Response of Spring–Mass System to an Exponential Decay

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