Question

The model ${\mathbf{mx}}^{\mathbf{"}}\mathbf{+}\mathbf{kx}\mathbf{+}{\mathbf{k}}_{\mathbf{1}}{\mathbf{x}}^{\mathbf{3}}\mathbf{=}{\mathbf{F}}_{\mathbf{0}}\mathbf{cos\omega t}$of an undamped periodically driven spring/mass system is called Duffing's differential equation. Consider the initial-value problem${\mathbf{x}}^{\mathbf{"}}\mathbf{+}\mathbf{x}\mathbf{+}{\mathbf{k}}_{\mathbf{1}}{\mathbf{x}}^{\mathbf{3}}\mathbf{=}\mathbf{5}\mathbf{cost}\mathbf{,}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$. Use a numerical solver to investigate the behavior of the system for values of ${\mathbf{k}}_{\mathbf{1}}\mathbf{>}\mathbf{0}$ranging from${\mathit{k}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$to${\mathit{k}}_{\mathbf{1}}\mathbf{=}\mathbf{100}$. State your conclusions.

Short answer

For larger value of ${\mathrm{K}}_1$, the nonlinearity effect $\left({\mathrm{k}}_1{\mathrm{x}}^3\right)$, which represents the restoring force of the spring, appears clearly in the solutions of the given initial value problem, maintaing the effect of increasing the amplitude of the oscillations due to the driven force. For very smaller values of ${\mathrm{K}}_1$, the nonlinearity effect becomes smaller, leading the driven force of the system to increase the amplitude of oscillations over time, which implies that the system approaches a pure resonance in that case. This analysis appears clearly at ${\mathrm{k}}_1=0$.

$\begin{array}{l}\ln \left[25\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+\mathrm{x}\text{'}\left[\mathrm{t}\right]=5\cos \left[\mathrm{t}\right],\mathrm{x}\left[0\right]=1,{\mathrm{x}}^{\text{'}}\left[0\right]=0\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,40\}]\\ \mathrm{Out}\left[25\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,40.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

$\mathrm{In}\left[33\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\right[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}25],\{\mathrm{t},0,40\},\mathrm{Plot}\mathrm{Legends}\to "{\mathrm{kk}}_1=0",\mathrm{PlotRange}\to \mathrm{Full}]$

Step-by-step solution

To Find the equation

We have

${\mathrm{x}}^{"}\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)+{\mathrm{k}}_1{\mathrm{x}}^3\left(\mathrm{t}\right)={\mathrm{F}}_0\mathrm{cost}$

Which represents a model for the undamped periodically spring/mass system, driven by the force${\mathrm{F}}_0$cos$\mathrm{t}$.By considering the initial value problem where

${\mathrm{x}}^{"}\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)+{\mathrm{k}}_1{\mathrm{x}}^3\left(\mathrm{t}\right)=5\mathrm{cost}\mathrm{and}\mathrm{x}\left(0\right)=1,{\mathrm{x}}^{\text{'}}\left(0\right)=0$

We solve the given initial value problem numerically, using Mathematica, by setting${\mathrm{k}}_1$to be any value between$0.01$ and 100 , yields

$\begin{array}{l}\ln \left[1\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+\mathrm{x}\text{'}\left[\mathrm{t}\right]+0.01\mathrm{x}{\left[\mathrm{t}\right]}^3=5\cos \left[\mathrm{t}\right],\mathrm{x}\left[0\right]=1,{\mathrm{x}}^{\text{'}}\left[0\right]=0\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,40\}]\\ \mathrm{Out}\left[1\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,40.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

$\begin{array}{l}\ln \left[2\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+\mathrm{x}\text{'}\left[\mathrm{t}\right]+0.05\mathrm{x}{\left[\mathrm{t}\right]}^3=5\cos \left[\mathrm{t}\right],\mathrm{x}\left[0\right]=1,{\mathrm{x}}^{\text{'}}\left[0\right]=0\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,40\}]\\ \mathrm{Out}\left[2\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,40.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

$\begin{array}{l}\ln \left[3\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+\mathrm{x}\text{'}\left[\mathrm{t}\right]+20\mathrm{x}{\left[\mathrm{t}\right]}^3=5\cos \left[\mathrm{t}\right],\mathrm{x}\left[0\right]=1,{\mathrm{x}}^{\text{'}}\left[0\right]=0\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,40\}]\\ \mathrm{Out}\left[3\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,40.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

$\begin{array}{l}\ln \left[4\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+\mathrm{x}\text{'}\left[\mathrm{t}\right]+50\mathrm{x}{\left[\mathrm{t}\right]}^3=5\cos \left[\mathrm{t}\right],\mathrm{x}\left[0\right]=1,{\mathrm{x}}^{\text{'}}\left[0\right]=0\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,40\}]\\ \mathrm{Out}\left[4\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,40.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

$\begin{array}{l}\ln \left[12\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+\mathrm{x}\left[\mathrm{t}\right]+100\mathrm{x}{\left[\mathrm{t}\right]}^3=5\cos \left[\mathrm{t}\right],\mathrm{x}\left[0\right]=1,{\mathrm{x}}^{\text{'}}\left[0\right]=0\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,40\}]\\ \mathrm{Out}\left[12\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{InterpolatingFunction}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,40.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

The two sets of the initial condition

Plotting the results, yields

$\mathrm{In}\left[11\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}1\right],\{\mathrm{t},0,40\},\mathrm{PlotLegends}\to "{\mathrm{k}}_1=0.{01}^{\mathrm{n}"}\right]$

$\mathrm{In}\left[13\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}2\right],\{\mathrm{t},0,40\},\mathrm{PlotLegends}\to "{\mathrm{k}}_1=0.{05}^{\mathrm{n}"}\right]$

$\mathrm{In}\left[15\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}3\right],\{\mathrm{t},0,40\},\mathrm{PlotLegends}\to "{\mathrm{k}}_1=20"\right]$

$\mathrm{In}\left[16\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}4\right],\{\mathrm{t},0,40\},\mathrm{PlotLegends}\to "{\mathrm{k}}_1=50"\right]$

$\mathrm{In}\left[23\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}12\right],\{\mathrm{t},0,40\},\mathrm{PlotLegends}\to "{\mathrm{k}}_1=100"\right]$

Conclusions

For larger value of ${\mathrm{k}}_1$, the nonlinearity effect $\left(k_1{x}^3\right)$, which represents the restoring force of the spring, appears clearly in the solutions of the given initial value problem, maintaing the effect of increasing the amplitude of the oscillations due to the driven force. For very smaller values of ${\mathrm{k}}_1$, the nonlinearity effect becomes smaller, leading the driven force of the system to increase the amplitude of oscillations over time, which implies that the system approaches a pure resonance in that case. This analysis appears clearly at $k_1=0$.

Final Answer

For larger value of ${\mathrm{k}}_1$, the nonlinearity effect $\left({\mathrm{k}}_1{\mathrm{x}}^3\right)$, which represents the restoring force of the spring, appears clearly in the solutions of the given initial value problem, maintaing the effect of increasing the amplitude of the oscillations due to the driven force. For very smaller values of ${\mathrm{k}}_1$, the nonlinearity effect becomes smaller, leading the driven force of the system to increase the amplitude of oscillations over time, which implies that the system approaches a pure resonance in that case. This analysis appears clearly at $k_1=0$.

$\begin{array}{l}\ln \left[25\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+\mathrm{x}\text{'}\left[\mathrm{t}\right]=5\cos \left[\mathrm{t}\right],\mathrm{x}\left[0\right]=1,{\mathrm{x}}^{\text{'}}\left[0\right]=0\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,40\}]\\ \mathrm{Out}\left[25\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{InterpolatingFunction}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,40.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

$\mathrm{In}\left[33\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\right[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}25],\{\mathrm{t},0,40\},\mathrm{PlotLegends}\to "{\mathrm{kk}}_1=0",\mathrm{PlotRange}\to \mathrm{Full}]$