Question

(a) Find numerical values of the constants and computer plot together on the same axis’s graphs of$(5.30),(5.31)$and$(\mathbf{5}.\mathbf{32})$ to compare overdamped, critically damped, and oscillatory motion. Suggested numbers: Let,$\omega =1$ and$b=13/5,1,5/13$for the three kinds of motion. Let $y\left(0\right)=1$&.$y^{\text{'}}\left(0\right)=0$

(b) Repeat the problem with the same set ofandvalues and with$y\left(0\right)=1$, but with.$y^{\text{'}}\left(0\right)=1$

(c) Again repeat, with.$y^{\text{'}}\left(0\right)=-1$

Short answer

(a) The overdamped case, $y=\frac{23}{24}{e}^{-5t}+\frac{1}{24}{e}^{-t/5}$the critically damped case $y=(1+t)e^{-t}$the oscillatory casedata-custom-editor="chemistry" $y=$$e^{-5t/13}(\frac{5}{12}\sin \frac{12}{13}t+\cos \frac{12}{13}t)$

(b) The overdamped case,$y=-\frac{1}{4}{e}^{-5t}+\frac{5}{4}{e}^{-t/5}$the critically damped case,$y=(1+2t)e^{-t}$ the oscillatory case$y=$ $e^{-5t/13}(\frac{6}{13}\sin \frac{12}{13}t+\cos \frac{12}{13}t)$

(c) The overdamped case,$y=-\frac{76}{24}{e}^{-5t}+\frac{100}{24}{e}^{-t/5}$ the critically damped case $y=e^{-t}$, the oscillatory case $y=e^{-5t/13}(-\frac{2}{3}$sin$\frac{12}{13}t+$ cos$\frac{12}{13}t)$.

Step-by-step solution

Step 1:Given information

Given is$\omega =1,b=\frac{13}{5},1,\frac{5}{13}$

Conditions for underdamped case, overdamped and for critical damped

The conditions for these cases are,

-For the overdamped case-, $b^2>{\omega }^2$and the solution for it is,$y=A{e}^{-\lambda t}+B{e}^{\mu t}$ where,$\mathit{\lambda }\mathbf{=}\mathit{b}\mathbf{+}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}{\mathbf{\omega }}^{\mathbf{2}}}$ and.$\mu =b-\sqrt{b^2-{\omega }^2}$

-For the critically damped case- the condition is$b^2={\omega }^2$ , and the solution is $y=(A+Bt)e^{-bt}$.

-For the underdamped (oscillatory)- case the condition is$b^2<{\omega }^2$ , and the solution is$y=e^{-bt}(Asin\beta t+Bcos\beta t)$ .

For the overdamped case-we will choose$\omega =1$ and.$b=13/5$ Therefore, the value of$\lambda =5$ & $\mu =1/5$.

Apply the initial condition

(a)

Find the first derivative to apply the initial condition to find the value of the constants in the solution,

$y^{\text{'}}=-\lambda A{e}^{-\lambda t}-\mu B{e}^{-\mu t}$

Now, we apply the initial conditions,

$\begin{array}{c}y\left(0\right)=1\\ =A+B\\ A=1-B\mathrm{............}\left(1\right)\end{array}$

$\begin{array}{c}{y}^{\text{'}}\left(0\right)=0\\ =-\lambda A-\mu B\mathrm{.......}\left(2\right)\end{array}$

Substitute $A$into eq.(2), so we get

$\begin{array}{l}-\lambda +\lambda B-\mu B\\ =0B(\lambda -\mu )\\ =\lambda B\\ =\frac{\lambda }{\lambda -\mu }\end{array}$

Solve further

$\begin{array}{l}=\frac{5}{5-1/5}\\ =\frac{1}{24}\end{array}$

therefore, using eq. (1)

$A=\frac{23}{24}$

So, the solution is

$y=\frac{23}{24}{e}^{-5t}+\frac{1}{24}{e}^{-t/5}$

Graph is

For the critically damped case

For the critically damped case we choose, $\omega =b=1$and we derive the solution to get

$y^{\text{'}}=B{e}^{-bt}-b(A+Bt)e^{-bt}$

Now, applying the boundary conditions, so we get

$\begin{array}{c}y\left(0\right)=1\\ =A\end{array}$

Now,

$\begin{array}{c}{y}^{\text{'}}\left(0\right)=0\\ =B-bA\\ B=1\end{array}$

So, the solution is

$y=(1+t)e^{-t}$

Graph is

 Step 5: For the oscillatory case

For the oscillatory case we choose, $\omega =1$and $b=5/13$, so the value of $\beta =12/13$, and the first derivative of the general solution is

$y^{\text{'}}=-b{e}^{-bt}(A\sin \beta t+B\cos \beta t)+e^{-bt}\beta (A\cos \beta t-\sin \beta t)$

Now, apply the initial conditions,

$\begin{array}{l}y\left(0\right)=1\\ =B{y}^{\text{'}}\left(0\right)\\ =0\\ =-bB+\beta AA\end{array}$

Solve further

$\begin{array}{l}=\frac{bB}{\beta }\\ =\frac{5}{12}\end{array}$

So, the solution now is

$y=e^{-5t/13}(\frac{5}{12}\sin \frac{12}{13}t+\cos \frac{12}{13}t)$

Graph is

Step 6:The combined graph of the cases

Combine the three, to get

(b) Step 7: For the overdamped case

This time the initial conditions are$y\left(0\right)=1$and $y^{\text{'}}\left(0\right)=1$. The constants will have the value as part). For the overdamped case (i.e. $\omega =1$,, and$b=13/5$) and have

$\begin{array}{c}y\left(0\right)=1\\ =A+B\\ =1\\ B=1-A\end{array}$

Now,

$\begin{array}{c}{y}^{\text{'}}\left(0\right)=1\\ =-5A-\frac{1}{5}\\ =-5A-\frac{1}{5}+\frac{1}{5}\end{array}$

Therefore,

$\begin{array}{c}A=1\\ A=-\frac{1}{4},\text{and}B=\frac{5}{4}\end{array}$

So, the solution is

$y=-\frac{1}{4}{e}^{-5t}+\frac{5}{4}{e}^{-t/5}$

Graph is

For the critical damped case

For the critical damped case i.e $(\omega =b=1)$.,we have

$\begin{array}{c}y\left(0\right)=1\\ =A\\ y^{\text{'}}\left(0\right)=1\\ =B-1\\ B=2\end{array}$

So, the solution becomes

$y=(1+2t)e^{-t}$

Graph is

For the oscillatory case

For the oscillatory case (i.e.,$\omega =1,and$) have

$\begin{array}{c}y\left(0\right)=1\\ =B{y}^{\text{'}}\\ =1\\ =-\frac{5}{13}+\frac{12}{13}A\end{array}$

Solve further

$A=\frac{6}{13}$

So, the solution is

$y=e^{-5t/13}(\frac{6}{13}\sin \frac{12}{13}t+\cos \frac{12}{13}t)$

The graph is

 Step 10: The combined graph

And when combine the three graphs to get.

(c) Step 11: For the overdamped case  

This time the initial conditions are $y\left(0\right)=1$and$y^{\text{'}}\left(0\right)=-1$the constants will have the value as part (a). For the overdamped case (i.e $\omega =1$.,, and$b=13/5$) have

$\begin{array}{c}y\left(0\right)=1=A+B\\ A=1-B\\ y^{\text{'}}\left(0\right)=-1\\ =-5+5B-\frac{1}{5}B\end{array}$

So, the solution is

$\therefore A=-\frac{76}{24},\text{and}B=\frac{100}{24}$

Graph is

For the critically damped case

For the critically damped case (i.e. $\omega =b=1$, ) have

$\begin{array}{c}y\left(0\right)=1=A\\ y^{\text{'}}\left(0\right)=-1\\ =B-1\\ B=0\end{array}$

So, the solution is

$y=e^{-t}$

Graph is

For the oscillatory case

For the oscillatory case (i.e.,$\omega =1,\&b=5/13$) and have

$\begin{array}{c}y\left(0\right)=1=B\\ y^{\text{'}}\left(0\right)=-1\\ =-\frac{5}{13}+\frac{12}{13}A\\ A=-\frac{2}{3}\end{array}$

So, the solution is

$y=e^{-5t/13}(-\frac{2}{3}\sin \frac{12}{13}t+\cos \frac{12}{13}t)$

Graph is

Step 14:The combined graph

Combine the graph three, and get