Q43E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A mass–spring system is driven by a sinusoidal external force $\mathbf{g}\left(\mathbf{t}\right)\mathbf{=}\mathbf{5}\mathbf{sint}$. The mass equals 1, the spring constant equals 3, and the damping coefficient equals 4. If the mass is initially located at  $\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\frac{1}{2}$and at rest, i.e., $\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, find its equation of motion.

Step-by-Step Solution

Answer

The equation of motion is $\mathrm{y}=2{\mathrm{e}}^{-\mathrm{t}}-\frac{1}{2}{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint}.$

Step 1: Use the given information for writing the differential equation.

Given that, 

 

The mass equals $\mathrm{m}=1$, 

 

The spring constant equals $\mathrm{c}=3$, 

 

And the damping coefficient equals $\mathrm{b}=4$.

 

The differential equation is,

 $\begin{array}{c}\mathrm{my}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\mathrm{g}\left(\mathrm{t}\right)\\ \mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+3\mathrm{y}=5\mathrm{sint}......\left(1\right)\end{array}$

 

Write the homogeneous differential equation of the equation (1),

 

$\mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+3\mathrm{y}=0$

Step 2: Now find the complementary solution of the given equation.

The auxiliary equation for the above equation,

 

${\mathrm{m}}^2+4\mathrm{m}+3=0$

 

Solve the auxiliary equation,

 

$\begin{array}{c}{\mathrm{m}}^2+4\mathrm{m}+3=0\\ {\mathrm{m}}^2+3\mathrm{m}+\mathrm{m}+3=0\\ \mathrm{m}(\mathrm{m}+3)+1(\mathrm{m}+3)=0\\ (\mathrm{m}+1)(\mathrm{m}+3)=0\end{array}$

 

The roots of auxiliary equation are, 

 ${\mathrm{m}}_1=-1,{\mathrm{m}}_2=-3$

The complimentary solution of the given equation is,

 ${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{t}}$

Step 3: Now find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acost}+\mathrm{Bsint} ......\left(2\right)$

Now find the first and second derivative of above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=-\mathrm{Asint}+\mathrm{Bcost}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=-\mathrm{Acost}-\mathrm{Bsint}\end{array}$

Substitute the value of  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$  and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$ in the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+3\mathrm{y}=5\mathrm{sint}\\ -\mathrm{Acost}-\mathrm{Bsint}+4(-\mathrm{Asint}+\mathrm{Bcost})+3(\mathrm{Acost}+\mathrm{Bsint})=5\mathrm{sint}\\ (2\mathrm{B}-4\mathrm{A})\mathrm{sint}+(2\mathrm{A}+4\mathrm{B})\mathrm{cost}=5\mathrm{sint}\end{array}$

Comparing the all coefficients of the above equation,

 $\begin{array}{c}2\mathrm{B}-4\mathrm{A}=5.......\left(3\right)\\ 4\mathrm{B}+2\mathrm{A}=0......\left(4\right)\end{array}$

Solve the above equations,

$\begin{array}{c}2(2\mathrm{B}-4\mathrm{A})=5\times 2\\ 4\mathrm{B}-8\mathrm{A}=10\\ 4\mathrm{B}+2\mathrm{A}=0\\ \mathrm{A}=-1\end{array}$

Substitute the value of A in the equation (4),

$\begin{array}{c}4\mathrm{B}+2(-1)=0\\ \mathrm{B}=\frac{1}{2}\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acost}+\mathrm{Bsint}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\mathrm{cost}+\frac{1}{2}\mathrm{sint}\end{array}$

Step 4: Find the general solution and use the given initial condition,

Therefore, the general solution is,

$\begin{array}{l}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint} ......\left(5\right)\end{array}$

Given initial condition,

$\mathrm{y}\left(0\right)=\frac{1}{2},\mathrm{y}\text{'}\left(0\right)=0$

Substitute the value of  $\mathrm{y}=\frac{1}{2}$ and t = 0 in the equation (3),

$\begin{array}{c}\frac{1}{2}={\mathrm{c}}_1{\mathrm{e}}^{-\left(0\right)}+{\mathrm{c}}_2{\mathrm{e}}^{-3\left(0\right)}-\cos \left(0\right)+\frac{1}{2}\sin \left(0\right)\\ \frac{1}{2}={\mathrm{c}}_1+{\mathrm{c}}_2-1\\ {\mathrm{c}}_1+{\mathrm{c}}_2=\frac{3}{2}.....\left(6\right)\end{array}$

Now find the derivative of above equation,

$\mathrm{y}\text{'}=-{\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}-3{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{t}}+\mathrm{sint}+\frac{1}{2}\mathrm{cost}$

Substitute the value of y’ = 0 and t = 0 in the above equation,

$\begin{array}{c}0=-{\mathrm{c}}_1{\mathrm{e}}^{-\left(0\right)}-3{\mathrm{c}}_2{\mathrm{e}}^{-3\left(0\right)}+\sin \left(0\right)+\frac{1}{2}\cos \left(0\right)\\ 0=-{\mathrm{c}}_1-3{\mathrm{c}}_2+\frac{1}{2}\\ {\mathrm{c}}_1+3{\mathrm{c}}_2=\frac{1}{2} ......\left(7\right)\end{array}$

Solve the (6) and (7) equations,

$\begin{array}{c}{\mathrm{c}}_1+{\mathrm{c}}_2=\frac{3}{2}\\ {\mathrm{c}}_1+3{\mathrm{c}}_2=\frac{1}{2}\\ {\mathrm{c}}_2=-\frac{1}{2}\end{array}$

Substitute the value of  ${\mathrm{c}}_2$ in the equation (6),

$\begin{array}{c}{\mathrm{c}}_1+{\mathrm{c}}_2=\frac{3}{2}\\ {\mathrm{c}}_1+\left(\frac{-1}{2}\right)=\frac{3}{2}\\ {\mathrm{c}}_1=2\end{array}$

Substitute the value of ${\mathrm{c}}_1$ and ${\mathrm{c}}_2$ in the equation (5),

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint}\\ \mathrm{y}=2{\mathrm{e}}^{-\mathrm{t}}-\frac{1}{2}{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint}\end{array}$

Thus, the equation of motion is:

$\mathrm{y}=2{\mathrm{e}}^{-\mathrm{t}}-\frac{1}{2}{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint}$