Benchmark solution
=====================


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.


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:

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

The natural frequency of this oscillator is:

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


Calculation of the transient response
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

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

:math:`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: :math:`\Omega \mathrm{=}1\mathit{rd}\mathrm{/}s` and :math:`F\mathrm{=}1N`.


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:

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

By applying a sine force :math:`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: :math:`\stackrel{ˆ}{x}(\Omega )\mathrm{=}\frac{F}{K\mathrm{-}{\Omega }^{2}M+jC}`

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