Question

Consider the model of an undamped nonlinear spring/mass system given by ${\mathit{x}}^{\mathbf{"}}\mathbf{+}\mathbf{8}\mathit{x}\mathbf{-}\mathbf{6}{\mathit{x}}^{\mathbf{3}}\mathbf{+}{\mathit{x}}^{\mathbf{5}}\mathbf{=}\mathbf{0}$. Use a numericalsolver to discuss the nature of the oscillations of the system corresponding to the initial conditions:

$\begin{array}{cc}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{;}& \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{2}}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{;}& \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{;}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}& \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\sqrt{\mathbf{2}}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\end{array}$

Short answer

The results were deduced.

Step-by-step solution

To Find the equation

We have the undamped nonlinear spring/mass, which is given by

$x^{"}+8x-6x^3+x^5=0$

with the initial boundary conditions

$x\left(0\right)=1,x^{\text{'}}\left(0\right)=1$and$x\left(0\right)=-2,x^{\text{'}}\left(0\right)=\frac{1}{2}$

$x\left(0\right)=\sqrt{2},x^{\text{'}}=1$and$x\left(0\right)=2,x^{\text{'}}\left(0\right)=\frac{1}{2}$

$x\left(0\right)=2,x^{\text{'}}\left(0\right)=0$and$x\left(0\right)=-\sqrt{2},x\text{'}\left(0\right)=-1$

Our aim is to solve, numerically, the given differential equation, subjected the given boundary conditions. Using Mathematical, we have

$\begin{array}{l}\mathrm{The}\mathrm{first}\mathrm{set}\mathrm{of}\mathrm{the}\mathrm{boundary}\mathrm{conditions}\\ \mathrm{We}\mathrm{have}\end{array}$

$\ln \left[14\right]:=\mathrm{NDSolve}[\left\{{\mathrm{x}}^{"}\left[\mathrm{t}\right]+8\mathrm{x}\left[\mathrm{t}\right]-6\mathrm{x}{\left[\mathrm{t}\right]}^3+\mathrm{x}{\left[\mathrm{t}\right]}^5=0,\mathrm{x}\left[0\right]=1,{\mathrm{x}}^{\text{'}}\left[0\right]=1\right\},\mathrm{x}\left[\mathrm{t}\right],\{\mathrm{t},0,14\}]$

$Out\left[14\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{InterpolatingFunction}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}$

$\begin{array}{l}\ln \left[15\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+8\mathrm{x}\left[\mathrm{t}\right]-6\mathrm{x}{\left[\mathrm{t}\right]}^3+\mathrm{x}{\left[\mathrm{t}\right]}^5=0,\mathrm{x}\left[0\right]=-2,{\mathrm{x}}^{\text{'}}\left[0\right]=\frac{1}{2}\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,14\}]\\ \mathrm{Out}\left[15\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{InterpolatingFunction}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

$\begin{array}{l}\mathrm{Plotting}\mathrm{the}\mathrm{results},\mathrm{yields}\\ \ln \left[17\right]:=\mathrm{Plot}[\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}14\right],\{\mathrm{t},0,14\},\mathrm{PlotLegends}{\to }^{"}\mathrm{x}(0)=1,\mathrm{x}\text{'}(0)=1^{"}]\end{array}$

$\ln \left[18\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\left[\mathrm{x}\right[\mathrm{t}\left]\right]/.\mathrm{\%}15],\{\mathrm{t},0,14\},\mathrm{PlotLegends}{\to }^{"}\mathrm{x}(0)=-2,\mathrm{x}\text{'}(0)={\frac{1}{2}}^{"}\right]$

For the first initial conditions, the period of oscillations is

$T\simeq 4.5-0.5=4sandA\simeq 1.2$

For the second initial conditions, we have

$\mathrm{T}\simeq 2.3-0.5=1.8\mathrm{s}\mathrm{and}\mathrm{A}\simeq |-1.85-(-2\left)\right|=0.15$

The two sets of the initial condition

The second set of the boundary conditions

We have

$\begin{array}{l}\ln \left[19\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+8\mathrm{x}\left[\mathrm{t}\right]-6\mathrm{x}{\left[\mathrm{t}\right]}^3+\mathrm{x}{\left[\mathrm{t}\right]}^5=0,\mathrm{x}\left[0\right]=\sqrt{2},{\mathrm{x}}^{\text{'}}\left[0\right]=1\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,14\}]\\ \mathrm{Out}\left[19\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{InterpolatingFunction}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

$\begin{array}{l}\ln \left[20\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+8\mathrm{x}\left[\mathrm{t}\right]-6\mathrm{x}{\left[\mathrm{t}\right]}^3+\mathrm{x}{\left[\mathrm{t}\right]}^5=0,\mathrm{x}\left[0\right]=2,{\mathrm{x}}^{\text{'}}\left[0\right]=\frac{1}{2}\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,14\}]\\ \mathrm{Out}\left[20\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\end{array}$

Plotting the results, yields

$\ln \left[23\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\right[\mathrm{x}\left[\mathrm{t}\right]]/.\mathrm{\%}19],\{\mathrm{t},0,14\},\mathrm{Plot}\mathrm{Legends}{\to }^{"}\mathrm{x}\left(0\right)=\sqrt{2},\mathrm{x}\text{'}\left(0\right)=1^{"}]$

$ln\left[24\right]:=Plot\left[Evaluate\right[x\left[t\right]]/.\%20],\{t,0,14\},PlotLegends{\to }^{"}x\left(0\right)=2,x\text{'}\left(0\right)={\frac{1}{2}}^{"}$

For the first initial conditions, the period of oscillations is

$\mathrm{T}\simeq 7-0.8=6.2\mathrm{s}\mathrm{and}\mathrm{A}\simeq 2.2$

For the second initial conditions, we have

$\mathrm{T}\simeq 2-0.3=1.7\mathrm{s}\mathrm{and}\mathrm{A}\simeq 2.11-2=0.11$

The third set of the boundary conditions

We have

$\begin{array}{l}\ln \left[4\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+8\mathrm{x}\left[\mathrm{t}\right]-6\mathrm{x}{\left[\mathrm{t}\right]}^3+\mathrm{x}{\left[\mathrm{t}\right]}^5=0,\mathrm{x}\left[0\right]=2,{\mathrm{x}}^{\text{'}}\left[0\right]=1\right\},\mathrm{x}\left[\mathrm{t}\right],\\ \{\mathrm{t},0,14\}]\end{array}$

$\mathrm{Out}\left[4\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}$

$\begin{array}{l}\ln \left[8\right]:=\mathrm{NDSolve}[\left\{\mathrm{x}"\left[\mathrm{t}\right]+8\mathrm{x}\left[\mathrm{t}\right]-6\mathrm{x}{\left[\mathrm{t}\right]}^3+\mathrm{x}{\left[\mathrm{t}\right]}^5=0,\mathrm{x}\left[0\right]=-\sqrt{2},{\mathrm{x}}^{\text{'}}\left[0\right]=-1\right\},\\ \mathrm{x}\left[\mathrm{t}\right],\{\mathrm{t},0,14\}]\end{array}$

$\mathrm{Out}\left[8\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}$

Plotting the results, yields

$\ln \left[9\right]:=\mathrm{Plot}\left[\mathrm{Evaluate}\right[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}4],\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}8\right],\mathrm{Evaluate}[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}8\left]\right\},$

$\{\mathrm{t},0,14\},\mathrm{PlotLegends}\to \left\{{\mathrm{n}}_{\mathrm{x}}\mathrm{x}\left(0\right)=2,{\mathrm{x}}^{\text{'}}\left(0\right)=0^{\mathrm{n}},{\mathrm{n}}^{\text{'}}\mathrm{x}\left(0\right)=-\sqrt{2},\mathrm{x}\text{'}\left(0\right)=-1^{\mathrm{n}}\right\}]$

Final Answer

For the first initial conditions, the period of oscillations is

$\mathrm{T}\approx 4-0.6=3.4\mathrm{s}\mathrm{and}\mathrm{A}\simeq 2.2$

For the second initial conditions, the solution is not an oscillatory.

See graphs and the results we deduced.