2. Benchmark solution

2.1. Calculation method

It is proposed to calculate the natural frequency of the oscillator, the response due to sinusoidal excitation and the response due to harmonic excitation.

2.1.1. Natural frequency calculation

To validate the natural frequency calculation, we consider the system without damping. Thus, the movement of the free end of the spring is governed by the following relationship:

\(M\ddot{x}+Kx\mathrm{=}F\) (1)

The natural frequency of this oscillator is:

\({f}_{0}\mathrm{=}\frac{1}{2\pi }\sqrt{\frac{K}{M}}\mathrm{=}\frac{{\omega }_{0}}{2\pi }\) (\({f}_{0}\): natural frequency, \({\omega }_{0}\): natural pulsation)

2.1.2. Calculation of the transient response

To validate the calculation of the transient response, we consider the system without damping.

With the initial condition \(x(0)\mathrm{=}0\) and \(\dot{x}(0)\mathrm{=}0\), if we apply a sinusoidal force \(F(t)\mathrm{=}F\mathrm{sin}(\Omega t)\), to the free end of the spring, the solution of the differential equation (1) is:

\(x(t)\mathrm{=}\frac{F(\mathrm{sin}\Omega t\mathrm{-}\frac{\Omega }{{\omega }_{0}}\mathrm{sin}{\omega }_{0}t)}{M({\omega }_{0}^{2}\mathrm{-}{\Omega }^{2})}\)

For this calculation of the response to a sinusoidal excitation, we chose: \(\Omega \mathrm{=}1\mathit{rd}\mathrm{/}s\) and \(F\mathrm{=}1N\).

2.1.3. Harmonic response calculation

It is then proposed to calculate the response of the damped oscillator due to harmonic excitation.

The movement of the free end of the spring is governed by the following relationship:

\(M\ddot{x}+C\dot{x}+Kx\mathrm{=}F\) (2)

By applying a sine force \(F(t)\mathrm{=}F\mathrm{sin}(\Omega t)\), to the free end of the spring, and by adopting the complex notation, we obtain the forced response: \(\stackrel{ˆ}{x}(\Omega )\mathrm{=}\frac{F}{K\mathrm{-}{\Omega }^{2}M+jC}\)

For this calculation of the harmonic forced response, we chose: \(0.5{\omega }_{0}\mathrm{\le }\Omega \mathrm{\le }1.5{\omega }_{0}\)