Question

A mass weighing 4 pounds is suspended from a spring whose constant is$\mathbf{3}\mathbf{\text{lb}}\mathbf{/}\mathbf{\text{ft}}\mathbf{.}$The entire system is immersed in a fluid offering a damping force numerically equal to the instantaneous velocity. Beginning at t=0, an external force equal to$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{t}}$is impressed on the system. Determine the equation of motion if the mass is initially released from rest at a point 2 feet below the equilibrium position.

Short answer

$x\left(t\right)=\frac{26}{17}{e}^{-4t}\cos 2\sqrt{2}t+\frac{28\sqrt{2}}{17}{e}^{-4t}\sin 2\sqrt{2}t+\frac{8}{17}{e}^{-t}$

Step-by-step solution

Step 1:Definition of Non-linear and Linear Spring

NONLINEAR SPRINGS The mathematical model has the form

$m\frac{d^{2}x}{d{t}^2}+F\left(x\right)=0$

where. Becausedenotes the displacement of the mass from its equilibrium position,$F\left(x\right)=kx$is Hooke's law-that is, the force exerted by the spring that tends to restore the mass to the equilibrium position. A spring acting under a linear restoring force$F\left(x\right)=kx$is naturally referred to as a linear spring.

A spring whose mathematical model incorporates a nonlinear restorative force, such as

$m\frac{d^{2}x}{d{t}^2}+k{x}^3=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}m\frac{d^{2}x}{d{t}^2}+kx+k_1{x}^3=0$

is called a nonlinear spring.

Find the auxiliary equation

We were given the next informations:

- A mass is weighing 4 pounds

- A spring constant is$3\text{lb}/\text{ft}$

- The damping force is equal to the instantaneous velocity

-$x\left(0\right)=2\text{ft}$

-$x^{\text{'}}\left(0\right)=0\text{ft}/\text{s}$

- The external force$f\left(t\right)=e^{-t}$is impressed on the system

We have

$W=mg\iff m=\frac{W}{g}=\frac{4\text{lb}}{32\text{ft}/{\text{s}}^2}=\frac{1}{8}\text{slug}$

From this, and the third and the sixth point of information’s, we have the equation that represents the motion:

$\frac{1}{8}{x}^{\text{'}\text{'}}+x^t+3x=e^{-t}$

The zeros of the auxiliary equation are

$x_{1,2}=\frac{-1\pm \sqrt{1-4\cdot \frac{1}{8}\cdot 3}}{\frac{1}{4}}\iff x_{1,2}=-4\pm 2i\sqrt{2}$

Therefore, the homogeneous solution of Eq. (1) is

$x_h\left(t\right)=c_1{e}^{-tt}\cos 2\sqrt{2}t+c_2{e}^{-1t}\sin 2\sqrt{2}t$

Find the particular solution

Let us assume that the particular solution is of the form

$x_p\left(t\right)=A{e}^{-t}$

Its derivations are

$x_p^{\text{'}}\left(t\right)=-A{e}^{-t},x_p^{\text{'}\text{'}}\left(t\right)=A{e}^{-t}$

Substitute it into (1):

$\frac{1}{8}A{e}^{-t}-A{e}^{-t}+3A{e}^{-t}=e^t\leftrightarrow A=\frac{8}{17}$

Therefore, the particular solution is

$x_p\left(t\right)=\frac{8}{17}{e}^{-t}$

The general solution is then:

$x\left(t\right)=x_h+x_p=c_1{e}^{-4t}\cos 2\sqrt{2}t+c_2{e}^{-4t}\sin 2\sqrt{2}t+\frac{8}{17}{e}^{-t}$

From the first initial condition, we get,$c_1+8/17=2$ that is$c_1=\frac{26}{17}$. Let us derivate (4):

$x^{\text{'}}\left(t\right)=e^{-4t}(-4c_1\cos 2\sqrt{2}t-4c_2\sin 2\sqrt{2}t-2\sqrt{2}{c}_1\sin 2\sqrt{2}t+2\sqrt{2}{c}_2\cos 2\sqrt{2}t)-\frac{8}{17}{e}^{-t}$

From the second initial condition, we get $0=-4c_1+2\sqrt{2}{c}_2-\frac{8}{17}\stackrel{c_1=26/17}{\iff }{c}_2=\frac{28\sqrt{2}}{17}$Substituting $c_1$and$c_2$ into (4) gives the solution:

$x\left(t\right)=\frac{26}{17}{e}^{-4t}\cos 2\sqrt{2}t+\frac{28\sqrt{2}}{17}{e}^{-4t}\sin 2\sqrt{2}t+\frac{8}{17}{e}^{-t}$