Question

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}+\frac{dx}{dt}+x-x^3=0 \\ x\left(0\right)=0,x^{\text{'}}\left(0\right)=\frac{3}{2};x\left(0\right)=-1,x^{\text{'}}\left(0\right)=1 \end{gathered}$$

Short answer

See graphs. By plotting F( x ), we found that the spring is considered to be a soft one, forcing the system not to oscillate

Step-by-step solution

To Find the equation

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

$\frac{d^{2}x}{d{t}^2}+\frac{dx}{dt}+x-x^3=0$

With the boundary conditions

$x\left(0\right)=0,x^{\text{'}}\left(0\right)=\frac{3}{2}andx\left(0\right)=-1,x^{\text{'}}\left(0\right)=1$

The given differential equation has the form

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}=-\frac{dx}{dt}-\left(x-x^3\right) \\ \frac{d^{2}x}{d{t}^2}=-\underset{Dampingterm}{\underbrace{\frac{dx}{dt}}}+F\left(x\right) \\ ma=\sum _i{F}_i[Newton\text{'}ssecondlaw] \end{gathered}$$

Where $\text{F}\left(\text{x}\right)$is the restoring force of the spring and it is equal to $-\left(x-x^3\right)$. And the negative sign means that the restoring force acts on a direction opposite to the direction of motion. Plotting the term $x-x^3$, yields

$\ln \left[30\right]:=Plot\left[x-x^3,\{x,-6,6\}\right]$

Which implies that the restoring force decreases as we we increase the elongation or compression of the spring, the restoring force has very low values corresponding to $x$ and the spring is considered, in that case, to be a soft one. We would not expect the system to have an oscillation behaviour in that case. For small elongation or compression, we might expect an oscillation behaviour. Now, we Solve numerically the given differential equation, using Mathematica, yields

$$\begin{gathered} ln\left[1\right]:=NDSolve\left[\left\{x^{\text{'}\text{'}}\left[t\right]+x\left[t\right]-x{\left[t\right]}^3=0,x\left[0\right]=0,x^{\text{'}}\left[0\right]=\frac{3}{2}\right\},x\left[t\right], \\ \left\{t,0,14\right\}\right] \\ out\left[1\right]=\left\{\left\{x\left[t\right]\to InterpolatingFunction\left[\begin{array}{c}Domain:\left\{\left\{0.,5.24\right\}\right\}\\ Output:scalar\end{array}\right]\left[t\right]\right\}\right\} \end{gathered}$$

$$\begin{gathered} ln\left[1\right]:=NDSolve\left[\left\{x^{\text{'}\text{'}}\left[t\right]+x\left[t\right]-x{\left[t\right]}^3=0,x\left[0\right]=0,x^{\text{'}}\left[0\right]=\frac{3}{2}\right\},x\left[t\right], \\ \left\{t,0,14\right\}\right] \\ out\left[1\right]=\left\{\left\{x\left[t\right]\to InterpolatingFunction\left[\begin{array}{c}Domain:\left\{\left\{0.,14.\right\}\right\}\\ Output:scalar\end{array}\right]\left[t\right]\right\}\right\} \end{gathered}$$

The two sets of the initial condition

Plotting the results, yields

$Plot\left[\right\{Evaluate\left[x\right[t]/.\%1],Evaluate\left[x\right[t]/.\%2]\},\{t,0,7\},$

$PlotLegends\to \left\{\left[n_x\left(0\right)=0,x^{\text{'}}\left(0\right)=\frac{3}{2},n_x\left(0\right)=-1,x^{\text{'}}\left(0\right)\right]=1^n\right\}$

Final Answer

Which testify our perdiction.

See graphs. By plotting F(x), we found that the spring is considered to be a soft one, forcing the system not to oscillate.