Q45E Speed Bumps. Often bumps like th... [FREE SOLUTION]

Chapter 4: Q45E (page 186)

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
$y (t) = 0$ for $t 鈮� \frac{- L}{2 V}$
localid="1655121580511" $my'' + ky = {\begin{cases} F_{0} \cos (\frac{蟺痴迟}{L}), 鈥夆赌� for 鈥� | t | < \frac{L}{2 V} \\ 0, 鈥夆赌夆€夆赌夆€� for 鈥� t 鈮� \frac{L}{2 V} \end{cases}}$
(The absence of a damping term indicates that the car鈥檚 shock absorbers are not functioning.)
Taking $m = k = 1, 鈥夆赌� L = 蟺$ , and $F_{0} = 1$ 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 $y (t) = Asint$ , where the constant A depends on the speed V.
Plot the amplitude |A| of the solution y(t) found in part (a) versus the car鈥檚 speed V. From the graph, estimate the speed that produces the most violent shaking of the vehicle.

Short Answer

Expert verified

The solution of the given problem is:

a. $y = \frac{2 V}{(1 - V^{2})} \cos (\frac{蟺}{2 V}) sint$

b. V = 0.73.

Step by step solution

Use the given information.

Given that,

$y (t) = 0$ for $t 鈮� \frac{- L}{2 V}$

And

$m = k = 1, 鈥夆赌� L = 蟺$ ,and $F_{0} = 1$

Consider the differential equation as:

role="math" localid="1655121904136" $my'' + ky = \{\begin{array}{l} F_{0} \cos (\frac{蟺 Vt}{L}), 鈥夆赌� for 鈥� | t | < \frac{L}{2 V} \\ 0, 鈥夆赌夆€夆赌夆€� for 鈥� t 鈮� \frac{L}{2 V} \end{array}\}$

Substitute the value of m, k, L, and $F_{0}$ in the above equation,

role="math" localid="1655122087786" $y'' + y = \{\begin{array}{l} \cos (Vt), 鈥夆赌� for 鈥� | t | < \frac{蟺}{2 V} \\ 0, 鈥夆赌夆€夆赌夆€� for 鈥� t 鈮� \frac{蟺}{2 V} \end{array}\} 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌� . ...... (1)$

Now find the solution of the given equation.

Given,

$y (t) = Asint$

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

$\begin{matrix} y' (t) = Acost \\ y'' (t) = - Asint \end{matrix}$

Now, one gets,

$y (t) + y'' (t) = Asint - Asint = 0$

Therefore, $y (t) = Asint$ is a solution for $t 鈮� \frac{蟺}{2 V}$ .

Solve for |t|<π2V and find the solution of the given equation.

Assume, the particular solution of equation (1),

$y_{p} (t) = c_{1} \cos (Vt) + c_{2} \sin (Vt) 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌� . ..... (2)$

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

$\begin{matrix} y_{p}' (t) = - c_{1} Vsin (Vt) + c_{2} Vcos (Vt) \\ y_{p}'' (t) = - c_{1} V^{2} \cos (Vt) - c_{2} V^{2} \sin (Vt) \end{matrix}$

We have;

role="math" localid="1655122388643" $y'' + y = \cos (Vt), 鈥夆赌� for 鈥� | t | < \frac{蟺}{2 V}$

Substitute the value of $y_{p} (t)$ and $y_{p}'' (t)$ in the above equation,

$\begin{matrix} y'' + y = \cos (Vt) \\ - c_{1} V^{2} \cos (Vt) - c_{2} V^{2} \sin (Vt) + c_{1} \cos (Vt) + c_{2} \sin (Vt) = \cos (Vt) \\ c_{1} (1 - V^{2}) \cos (Vt) + c_{2} (1 - V^{2}) \sin (Vt) = \cos (Vt) \end{matrix}$

Compare the coefficient of the above equation,

role="math" localid="1655122528934" $\begin{matrix} c_{1} (1 - V^{2}) = 1 \\ c_{1} = \frac{1}{(1 - V^{2})} \\ c_{2} (1 - V^{2}) = 0 \\ c_{2} = 0 \end{matrix}$

Substitute the value of $C_{1}$ and $C_{2}$ in the equation (2),

role="math" localid="1655122610131" $\begin{matrix} y_{p} (t) = \frac{1}{(1 - V^{2})} \cos (Vt) + (0) \sin (Vt) \\ y_{p} (t) = \frac{1}{(1 - V^{2})} \cos (Vt) \end{matrix}$

Therefore, the particular solution of equation (1),

role="math" localid="1655122652728" $y_{p} (t) = \frac{1}{(1 - V^{2})} \cos (Vt)$

And

$y_{c} (t) = Ccost + Dsint$

Find the general solution.

Therefore, the general solution is,

$\begin{matrix} y = y_{c} (t) + y_{p} (t) \\ y = Ccost + Dsint + \frac{1}{(1 - V^{2})} \cos (Vt) 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€� . ...... (3) \end{matrix}$

Check for $y (\frac{- 蟺}{2 V}) = 0$ ,

From the equation (3),

role="math" localid="1655124263786" $\begin{matrix} y (\frac{- 蟺}{2 V}) = 0 \\ Ccos (\frac{- 蟺}{2 V}) + Dsin (\frac{- 蟺}{2 V}) + \frac{1}{(1 - V^{2})} \cos (V (\frac{- 蟺}{2 V})) = 0 \\ Ccos (\frac{蟺}{2 V}) - Dsin (\frac{蟺}{2 V}) = 0 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌� . ..... (4) \end{matrix}$

We know that,

$y (t) = Asint$ for role="math" localid="1655123238268" $t 鈮� \frac{蟺}{2 V}$ .

Check for role="math" localid="1655123341011" $y (\frac{A}{2 V}) = Asin (\frac{A}{2 V})$ ,

From the equation (3),

role="math" localid="1655123857333" $\begin{matrix} y (\frac{蟺}{2 V}) = Asin (\frac{蟺}{2 V}) \\ Ccos (\frac{蟺}{2 V}) + Dsin (\frac{蟺}{2 V}) + \frac{1}{(1 - V^{2})} \cos (V \frac{蟺}{2 V}) = Asin (\frac{蟺}{2 V}) \\ Ccos (\frac{蟺}{2 V}) + Dsin (\frac{蟺}{2 V}) = Asin (\frac{蟺}{2 V}) 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€� . ..... (5) \end{matrix}$

We can write as:

role="math" localid="1655124056135" $y' (\frac{蟺}{2 V}) = Acos (\frac{蟺}{2 V})$

From the equation (3),

role="math" localid="1655124329438" $\begin{matrix} y (\frac{蟺}{2 V}) = Acos (\frac{蟺}{2 V}) \\ - Csin (\frac{蟺}{2 V}) + Dcos (\frac{蟺}{2 V}) - \frac{V}{(1 - V^{2})} \sin (V (\frac{蟺}{2 V})) = Acos (\frac{蟺}{2 V}) \\ - Csin (\frac{蟺}{2 V}) + Dcos (\frac{蟺}{2 V}) + \frac{V}{(V^{2} - 1)} = Acos (\frac{蟺}{2 V}) 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€� . ..... (6) \end{matrix}$

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

role="math" localid="1655124410214" $Ccos (\frac{蟺}{2 V}) = Dsin (\frac{蟺}{2 V})$

And

\begin{matrix} 2 Dsin (\frac{蟺}{2 V}) = Asin (\frac{蟺}{2 V}) \\ A = 2 D \\ 2 D = \frac{2 V}{(1 - V^{2})} \cos (\frac{蟺}{2 V}) \\ D = \frac{V}{(1 - V^{2})} \cos (\frac{蟺}{2 V}) \end{matrix}

$y = Asint = \frac{2 V}{(1 - V^{2})} \cos (\frac{蟺}{2 V}) sint$

Use the given information.

$| A | = | \frac{2 V}{(1 - V^{2})} \cos (\frac{蟺}{2 V}) |$

Given, violent shaking occurs for V = 0.73.

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Short Answer

Step by step solution

Use the given information.

Now find the solution of the given equation.

Solve for |t|<π2V and find the solution of the given equation.

Find the general solution.

Use the given information.

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