Q45E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Speed Bumps. Often bumps like the one depicted in Figure 4.11 are built into roads to discourage speeding. The figure suggests that a crude model of the vertical motion y(t) of a car encountering the speed bump with the speed V is given by

  $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{0}$ for $\mathbf{t}\le \frac{-L}{2V}$

$\mathbf{my}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\mathbf{=}\mathbf{\left\{}\begin{array}{l}{\mathbf{F}}_{\mathbf{0}}\mathbf{cos}\left(\frac{\mathrm{\pi Vt}}{L}\right)\mathbf{,}\mathbf{for}\mathbf{\left|}\mathbf{t}\mathbf{\right|}\mathbf{<}\frac{\mathbf{L}}{\mathbf{2}\mathbf{V}}\\ \mathbf{0}\mathbf{,}\mathbf{for}\mathbf{t}\mathbf{\ge }\frac{\mathbf{L}}{\mathbf{2}\mathbf{V}}\end{array}\mathbf{\right\}}$

 

(The absence of a damping term indicates that the car’s shock absorbers are not functioning.) 

1. Taking  $\mathbf{m}\mathbf{=}\mathbf{k}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{L}\mathbf{=}\mathbf{\pi }$, and ${\mathbf{F}}_0\mathbf{=}\mathbf{1}$   in appropriate units, solve this initial value problem. Thereby showing that the formula for the oscillatory motion after the car has traversed the speed bump is  $\mathbf{y}\left(\mathbf{t}\right)=\mathbf{Asint}$, where the constant A depends on the speed V. 
2. Plot the amplitude |A| of the solution y(t) found in part (a) versus the car’s speed V. From the graph, estimate the speed that produces the most violent shaking of the vehicle.

Step-by-Step Solution

Answer

The solution of the given problem is:

a.$\mathrm{y}=\frac{2\mathrm{V}}{(1-{\mathrm{V}}^2)}\cos \left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\mathrm{sint}$

b. V = 0.73.

Step 1: Use the given information.

Given that, 

$\mathrm{y}\left(\mathrm{t}\right)=0$ for  $\mathbf{t}\le \frac{-L}{2V}$

And 

 

$\mathrm{m}=\mathrm{k}=1,\mathrm{L}=\mathrm{\pi }$ , and  ${\mathrm{F}}_0=1$

 

Consider the differential equation as:

  $\mathrm{my}\text{'}\text{'}+\mathrm{ky}=\left\{\begin{array}{l}{\mathrm{F}}_0\cos \left(\frac{\mathrm{\pi } \mathrm{Vt}}{\mathrm{L}}\right),\mathrm{for}\left|\mathrm{t}\right|<\frac{\mathrm{L}}{2\mathrm{V}}\\ 0,\mathrm{for}\mathrm{t}\ge \frac{\mathrm{L}}{2\mathrm{V}}\end{array}\right\}$

Substitute the value of m, k, L, and  ${\mathrm{F}}_0$  in the above equation,

$\mathrm{y}\text{'}\text{'}+\mathrm{y}=\left\{\begin{array}{l}\cos \left(\mathrm{Vt}\right),\mathrm{for}\left|\mathrm{t}\right|<\frac{\mathrm{\pi }}{2\mathrm{V}}\\ 0,\mathrm{for}\mathrm{t}\ge \frac{\mathrm{\pi }}{2\mathrm{V}}\end{array}\right\} .......\left(1\right)$

 

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

Given, 

 

 $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asint}$

 

Now find the first and second derivatives of the above equation,

 

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

 

Now, one gets,

 

 $\mathrm{y}\left(\mathrm{t}\right)+\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)=\mathrm{Asint}-\mathrm{Asint}=0$

Therefore, $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asint}$   is a solution for $\mathbf{t}\ge \frac{\pi }{2V}$ .

Step 3: Solve for | t | &#60; &#960; 2 V and find the solution of the given equation.

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(\mathrm{Vt}\right)+{\mathrm{c}}_2\sin \left(\mathrm{Vt}\right)......\left(2\right)$

Now find the first and second derivatives of the above equation,

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_1\mathrm{Vsin}\left(\mathrm{Vt}\right)+{\mathrm{c}}_2\mathrm{Vcos}\left(\mathrm{Vt}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_1{\mathrm{V}}^2\cos \left(\mathrm{Vt}\right)-{\mathrm{c}}_2{\mathrm{V}}^2\sin \left(\mathrm{Vt}\right)\end{array}$

We have;

$\mathrm{y}\text{'}\text{'}+\mathrm{y}=\cos \left(\mathrm{Vt}\right),\mathrm{for}\left|\mathrm{t}\right|<\frac{\mathrm{\pi }}{2\mathrm{V}}$

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

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+\mathrm{y}=\cos \left(\mathrm{Vt}\right)\\ -{\mathrm{c}}_1{\mathrm{V}}^2\cos \left(\mathrm{Vt}\right)-{\mathrm{c}}_2{\mathrm{V}}^2\sin \left(\mathrm{Vt}\right)+{\mathrm{c}}_1\cos \left(\mathrm{Vt}\right)+{\mathrm{c}}_2\sin \left(\mathrm{Vt}\right)=\cos \left(\mathrm{Vt}\right)\\ {\mathrm{c}}_1(1-{\mathrm{V}}^2)\cos \left(\mathrm{Vt}\right)+{\mathrm{c}}_2(1-{\mathrm{V}}^2)\sin \left(\mathrm{Vt}\right)=\cos \left(\mathrm{Vt}\right)\end{array}$

Compare the coefficient of the above equation,

  $\begin{array}{c}{\mathrm{c}}_1(1-{\mathrm{V}}^2)=1\\ {\mathrm{c}}_1=\frac{1}{(1-{\mathrm{V}}^2)}\\ {\mathrm{c}}_2(1-{\mathrm{V}}^2)=0\\ {\mathrm{c}}_2=0\end{array}$

Substitute the value of  ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$  in the equation (2),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{(1-{\mathrm{V}}^2)}\cos \left(\mathrm{Vt}\right)+\left(0\right)\sin \left(\mathrm{Vt}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{(1-{\mathrm{V}}^2)}\cos \left(\mathrm{Vt}\right)\end{array}$

Therefore, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{(1-{\mathrm{V}}^2)}\cos \left(\mathrm{Vt}\right)$

And 

${\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)=\mathrm{Ccost}+\mathrm{Dsint}$

Step 4: Find the general solution.

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}=\mathrm{Ccost}+\mathrm{Dsint}+\frac{1}{(1-{\mathrm{V}}^2)}\cos \left(\mathrm{Vt}\right).......\left(3\right)\end{array}$

Check for $\mathrm{y}\left(\frac{-\mathrm{\pi }}{2\mathrm{V}}\right)=0$ ,

 

From the equation (3),

$\begin{array}{c}\mathrm{y}\left(\frac{-\mathrm{\pi }}{2\mathrm{V}}\right)=0\\ \mathrm{Ccos}\left(\frac{-\mathrm{\pi }}{2\mathrm{V}}\right)+\mathrm{Dsin}\left(\frac{-\mathrm{\pi }}{2\mathrm{V}}\right)+\frac{1}{(1-{\mathrm{V}}^2)}\cos \left(\mathrm{V}\left(\frac{-\mathrm{\pi }}{2\mathrm{V}}\right)\right)=0\\ \mathrm{Ccos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)-\mathrm{Dsin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)=0......\left(4\right)\end{array}$

We know that,

$\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asint}$ for   $\mathrm{t}\ge \frac{\mathrm{\pi }}{2\mathrm{V}}$ .

Check for   $\mathrm{y}\left(\frac{\mathrm{A}}{2\mathrm{V}}\right)=\mathrm{Asin}\left(\frac{\mathrm{A}}{2\mathrm{V}}\right)$ ,

From the equation (3),

$\begin{array}{c}\mathrm{y}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)=\mathrm{Asin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\\ \mathrm{Ccos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)+\mathrm{Dsin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)+\frac{1}{(1-{\mathrm{V}}^2)}\cos \left(\mathrm{V}\frac{\mathrm{\pi }}{2\mathrm{V}}\right)=\mathrm{Asin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\\ \mathrm{Ccos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)+\mathrm{Dsin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)=\mathrm{Asin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right) ......\left(5\right)\end{array}$

We can write as:

$\mathrm{y}\text{'}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)=\mathrm{Acos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)$

From the equation (3),

$\begin{array}{c}\mathrm{y}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)=\mathrm{Acos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\\ -\mathrm{Csin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)+\mathrm{Dcos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)-\frac{\mathrm{V}}{(1-{\mathrm{V}}^2)}\sin \left(\mathrm{V}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\right)=\mathrm{Acos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\\ -\mathrm{Csin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)+\mathrm{Dcos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)+\frac{\mathrm{V}}{({\mathrm{V}}^2-1)}=\mathrm{Acos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)......\left(6\right)\end{array}$

Solve the equation (4), (5), and (6),

$\mathrm{Ccos}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)=\mathrm{Dsin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)$

And 

 

$\begin{array}{c}2\mathrm{Dsin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)=\mathrm{Asin}\left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\\ \mathrm{A}=2\mathrm{D}\\ 2\mathrm{D}=\frac{2\mathrm{V}}{(1-{\mathrm{V}}^2)}\cos \left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\\ \mathrm{D}=\frac{\mathrm{V}}{(1-{\mathrm{V}}^2)}\cos \left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\end{array}$

$\mathrm{y}=\mathrm{Asint}=\frac{2\mathrm{V}}{(1-{\mathrm{V}}^2)}\cos \left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\mathrm{sint}$

Step 5: Use the given information.

$\left|\mathrm{A}\right|=\left|\frac{2\mathrm{V}}{(1-{\mathrm{V}}^2)}\cos \left(\frac{\mathrm{\pi }}{2\mathrm{V}}\right)\right|$

Given, violent shaking occurs for V = 0.73.