Question

In the following problems, take$\mathrm{g}=32\mathrm{ft}/{\sec }^2$ for the U.S. Customary System and $\mathrm{g}=9.8\mathrm{m}/{\sec }^2$for the MKS system.

The response of an overdamped system to a constant force is governed by equation (1) with m = 2, b = 8, k = 6, and$F_o=18and\gamma =0$ . If the system starts from rest, $\left[y\left(0\right)=y\text{'}\left(0\right)=0\right],$compute and sketch the displacement y(t). What is the limit of y(t) as $t\to +\infty$? Interpret this physically.

Short answer

Therefore, the general solution is $y\left(t\right)=-\frac{9}{2}{e}^d+\frac{3}{2}{e}^a+3$and its sketch is shown below.

The physical interpretation of displacement will remain at 3 after some time.

Step-by-step solution

General form

The general solution to (1) in the case $0<b^2<4\mathrm{k}$:

$y\left(t\right)=A{e}^{-\left(\frac{b}{2m}\right)t}\sin \left(\frac{\sqrt{4mk-b^2}}{2m}t+\varnothing \right)+\frac{F_0}{\sqrt{{\left(k-m{y}^2\right)}^2}+b^2{y}^2}\sin \left(yt+\theta \right)$

The angular frequency:

The amplitude of the steady-state solution to equation (1) depends on the angular frequency $\gamma$of the forcing function and it is given by $\mathrm{A}\left(\gamma \right)={\mathrm{F}}_0\mathrm{M}\left(\gamma \right)$, were

$\mathrm{M}\left(\gamma \right):=\frac{1}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\gamma }^2\right)}^2+{\mathrm{b}}^2{\gamma }^2}} \dots \left(1\right)a$

The undamped system:

The system is governed by $\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{ky}={\mathrm{F}}_0\cos \gamma \mathrm{t}$. And the homogenous solution of it is ${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right),\omega :=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$. And the corresponding homogeneous equation is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$.

So, the general solution to the system is $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$.

Evaluate the equation

Referring to problem 7: one gets,

The homogeneous solution to $m\frac{d^{2}y}{d{t}^2}+b\frac{dy}{dt}+ky=F_0\cos \left(\gamma t\right)$and it becomes

$$\begin{gathered} y_h\left(t\right)=c_1{e}^{r_{1}t}+c_2{e}^{r_{2}t} \\ {y}_p\left(t\right)=\frac{F_0}{{\left(k-m{\gamma }^2\right)}^2+b^2{\gamma }^2}\left[\left(k-m{\gamma }^2\right)\cos \left(\gamma t\right)+b\gamma \sin \left(\gamma t\right)\right] \end{gathered}$$

Then, the general solution is

$y\left(t\right)=c_1{e}^{\left(-\frac{b}{2m}+\frac{1}{2m}\sqrt{b^2-4mk}\right)t}+c_2{e}^{\left(-\frac{b}{2m}-\frac{1}{2m}\sqrt{b^2-4mk}\right)t}+\frac{F_0}{{\left(k-m{\gamma }^2\right)}^2+b^2{\gamma }^2}\left[\left(k-m{\gamma }^2\right)\cos \left(\gamma t\right)+b\gamma \sin \left(\gamma t\right)\right] ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Given that, $m=2,b=8,k=6,F_0=18and\gamma =0$

Then, the equation is$2\frac{d^{2}y}{d{t}^2}+8\frac{dy}{dt}+6y=18$

Substitute the values of m, k, b, … in equation (2)

$$\begin{gathered} \left(t\right)=c_1{e}^{{\left(\frac{8}{2\times 2}+\frac{1}{2\times 2}\sqrt{8^2-4\times 2\times 6}\right)}^t}+c_2{e}^{{\left(\frac{8}{2\times 2}-\frac{1}{2\times 2}\sqrt{8^2-4\times 2\times 6}\right)}^t}+\frac{18}{{\left(6-0\right)}^2+0}\left[\left(6-0\right)\cos \left(0\right)+b\left(0\right)\sin \left(0\right)\right] \\ =c_1{e}^{-t}+c_2{e}^{-3t}+3 \end{gathered}$$

So, the solution is $y\left(t\right)=c_1{e}^{-1}+c_2{e}^{-3t}+3$.

Now find the derivative of y.

$y\text{'}\left(t\right)=-c_1{e}^{-t}-3c_2{e}^{-3t}$

Implement the initial conditions.

Given,$y\left(0\right)=y\text{'}\left(0\right)=0$

Then, substitute it y(t) and y’(t) to get the value of c’s.

$$\begin{gathered} \left(t\right)=c_1{e}^{-t}+c_2{e}^{-3t}+3 \\ y\left(0\right)=c_1{e}^{-0}+c_2{e}^{-3\left(0\right)}+3 \\ 0=c_1+c_2+3 \\ {c}_1+c_2=-3 \end{gathered}$$

And

$$\begin{gathered} \left(t\right)=-c_1{e}^{-t}-3c_2{e}^{-3t} \\ y\text{'}\left(0\right)=-c_1{e}^{-\left(0\right)}-3c_2{e}^{-3\left(0\right)} \\ 0=-c_1-3c_2 \\ {c}_1=-3c_2 \end{gathered}$$

Solve the above equations.

$$\begin{gathered} -3c_2+c_2=-3 \\ -2c_2=-3 \\ {c}_2=\frac{3}{2} \\ {c}_1=-3\left(\frac{3}{2}\right) \\ =-\frac{9}{2} \end{gathered}$$

Then, the solution becomes $y\left(t\right)=-\frac{9}{2}{e}^{-t}+\frac{3}{2}{e}^{-3t}+3$.

Sketch the graph of the equation and find its limit

The graph of the equation $y\left(t\right)=-\frac{9}{2}{e}^{-t}+\frac{3}{2}{e}^{-3t}+3$is shown below.

Since $e^{-t}and{e}^{-3t}$approaches zero. Then, the limit value must be zero.

That is,

$$\begin{gathered} \underset{t\to \infty }{\lim y\left(t\right)}=\underset{t\to \infty }{\lim }-\frac{9}{2}{e}^{-t}+\frac{3}{2}{e}^{-3t}+3 \\ =3 \end{gathered}$$

Then, physically this means that under a constant force, the displacement will pretty much remain at 3 after some time.