# Peak acceleration of a harmonic oscillator with a sinusoidal driving force

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

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