Peak acceleration of a harmonic oscillator with a sinusoidal driving force

Figure 1: Illustration of system
Figure 1: Illustration of system

At the natural frequency, for a lightly damped system (\(\zeta \ll 1\)), the peak acceleration is given by the expression

\[ a_{peak} = \frac{F_{0}}{2 \zeta m} \]

where

\[ \zeta = \frac{c}{2 \sqrt{mk}} \]

Derivation

The system in Figure 1 is described by the equation

\[\begin{align} \sum F &= ma \\ F_{0} \sin \omega t - c \frac{dx}{dt} - kx &= m \frac{d^{2}x}{dt^{2}} \\ \frac{d^{2}x}{dt^{2}} + \frac{c}{m}\frac{dx}{dt} + \frac{k}{m} x &= \frac{F_{0}}{m} \sin \omega t \end{align}\]

which can be expressed as

\[ \frac{d^{2}x}{dt^{2}} + 2 \zeta \omega_{n} \frac{dx}{dt} + w_{n}^{2} x = \frac{F_{0}}{m}\sin \omega t \]

where

\[\begin{align} \omega_{n} & = \sqrt{\frac{k}{m}} \\ \zeta & = \frac{c}{2 \sqrt{mk}} \end{align}\]

The solution to this equation expressed in terms of \(x\) is

\[ x(t) = \frac{F_{0}}{m Z_{m} \omega} \sin (\omega t + \varphi) \label{a}\tag{Eq.1} \]

where

\[\begin{align} Z_{m} & = \sqrt{(2 \omega_{n} \zeta)^{2} + \frac{1}{\omega^{2}}(\omega_{n}^{2} - \omega^{2})^{2}} \label{zm}\tag{Eq.2}\\ \varphi & = \tan^{-1} \Bigg( \frac{2 \omega \omega_{n} \zeta}{\omega^{2} - \omega_{n}^{2}} \Bigg) \end{align}\]

From \(\ref{a}\) we can write expressions for velocity and acceleration

\[\begin{align} \dot x (t) & = \frac{F_{0}}{m Z_{m}} \cos (\omega t + \varphi) \\ \ddot x (t) & = - \frac{F_{0} \omega}{m Z_{m}} \sin (\omega t + \varphi) \label{b}\tag{Eq.3} \end{align}\]

From \(\ref{b}\) we can write an expression for peak acceleration

\[ a_{peak} = \frac{F_{0} \omega}{m Z_{m}} \label{c}\tag{Eq.4} \]

We note that \(\ref{c}\) is at a maximum where \(\omega = \omega_{n} (1 - 2 \zeta^{2})^{-0.5}\) (see derivation), which we will approximate as \(\omega \approx \omega_{n}\) for small \(\zeta\) using a binomial approximation, which we will substitute into our expression for \(Z_{m}\) in \(\ref{zm}\) to give:

\[ Z_{m} = 2 \omega_{n} \zeta \label{d}\tag{Eq.5} \]

Substituting \(\ref{d}\) into \(\ref{c}\) we obtain

\[ a_{peak} = \frac{F_{0}}{2 \zeta m} \]