Chapter 19.5, Problem 19.127P

(a)

To determine

Show that in case of heavy damping $\left(c>c_c\right)$ a body never passes through its position of equilibrium O if it is released with no initial velocity from an arbitrary position.

Explanation of Solution

Calculation:

Since $\left(c>c_c\right)$ which is over damped, the wavelength ${\lambda }_1$ and ${\lambda }_2$ is less than zero.

The expression for the differential equation of over damping as follows:

$x=C_1{e}^{{\lambda }_{1}t}+C_2{e}^{{\lambda }_{2}t}$ (1)

Differentiate the above equation with respect to ‘t’.

$v=\frac{dx}{dt}=C_1{\lambda }_1{e}^{{\lambda }_{1}t}+C_2{\lambda }_2{e}^{{\lambda }_{2}t}$ (2)

Since the body is released with no initial velocity.

Substitute 0 for t, $x_0$ for x and 0 for v in the equation (1).

$\begin{array}{l}{x}_0=C_1{e}^{{\lambda }_1\left(0\right)}+C_2{e}^{{\lambda }_2\left(0\right)}\\ x_0=C_1{e}^0+C_2{e}^0\\ x_0=C_1\left(1\right)+C_2\left(1\right)\\ x_0=C_1+C_2\\ C_1=x_0-C_2\end{array}$ (3)

Substitute 0 for t, $x_0$ for x and 0 for v in the equation (2).

$\begin{array}{l}\frac{dx}{dt}=C_1{\lambda }_1{e}^{{\lambda }_{1}t}+C_2{\lambda }_2{e}^{{\lambda }_{2}t}\\ 0=C_1{\lambda }_1{e}^{{\lambda }_1\left(0\right)}+C_2{\lambda }_2{e}^{{\lambda }_2\left(0\right)}\\ 0=C_1{\lambda }_1\left(1\right)+C_2{\lambda }_2\left(1\right)\\ 0=C_1{\lambda }_1+C_2{\lambda }_2\end{array}$

Substitute $x_0-C_2$ for $C_1$.

$\begin{array}{l}{C}_1{\lambda }_1+C_2{\lambda }_2=0\\ \left(x_0-C_2\right){\lambda }_1+C_2{\lambda }_2=0\\ x_0{\lambda }_1-C_2{\lambda }_1+C_2{\lambda }_2=0\\ x_0{\lambda }_1-C_2\left({\lambda }_1-{\lambda }_2\right)=0\\ x_0{\lambda }_1=C_2\left({\lambda }_1-{\lambda }_2\right)\\ C_2\left({\lambda }_2-{\lambda }_1\right)=-x_0{\lambda }_1\\ C_2=\frac{-x_0{\lambda }_1}{\left({\lambda }_2-{\lambda }_1\right)}\end{array}$

Substitute $\frac{-x_0{\lambda }_1}{\left({\lambda }_2-{\lambda }_1\right)}$ for $C_2$ in the equation (3).

$\begin{array}{c}{C}_1=x_0-\frac{-x_0{\lambda }_1}{\left({\lambda }_2-{\lambda }_1\right)}\\ =x_0+\frac{x_0{\lambda }_1}{\left({\lambda }_2-{\lambda }_1\right)}\\ =\frac{x_0\left({\lambda }_2-{\lambda }_1\right)+x_0{\lambda }_1}{{\lambda }_2-{\lambda }_1}\\ =\frac{x_0{\lambda }_2-x_0{\lambda }_1+x_0{\lambda }_1}{{\lambda }_2-{\lambda }_1}\\ =\frac{x_0{\lambda }_2}{{\lambda }_2-{\lambda }_1}\end{array}$

Substitute $\frac{-x_0{\lambda }_1}{\left({\lambda }_2-{\lambda }_1\right)}$ for $C_2$ and $\frac{x_0{\lambda }_2}{{\lambda }_2-{\lambda }_1}$ for $C_1$ in equation (1).

$\begin{array}{c}x=\frac{x_0{\lambda }_2}{{\lambda }_2-{\lambda }_1}{e}^{{\lambda }_{1}t}-\frac{x_0{\lambda }_1}{\left({\lambda }_2-{\lambda }_1\right)}{e}^{{\lambda }_{2}t}\\ =\frac{x_0}{{\lambda }_2-{\lambda }_1}\left[{\lambda }_2{e}^{{\lambda }_{1}t}-{\lambda }_1{e}^{{\lambda }_{2}t}\right]\end{array}$

Apply boundary condition.

For $x=0$ when $t\ne \infty$.

$\begin{array}{l}0=\frac{x_0}{{\lambda }_2-{\lambda }_1}\left[{\lambda }_2{e}^{{\lambda }_{1}t}-{\lambda }_1{e}^{{\lambda }_{2}t}\right]\\ {\lambda }_2{e}^{{\lambda }_{1}t}-{\lambda }_1{e}^{{\lambda }_{2}t}=0\\ {\lambda }_2{e}^{{\lambda }_{1}t}={\lambda }_1{e}^{{\lambda }_{2}t}\\ \frac{{\lambda }_2}{{\lambda }_1}=\frac{e^{{\lambda }_{2}t}}{e^{{\lambda }_{1}t}}\\ \frac{{\lambda }_2}{{\lambda }_1}=e^{{\lambda }_{2}t}-e^{-{\lambda }_{1}t}\\ \frac{{\lambda }_2}{{\lambda }_1}=e^{{}^{{\lambda }_{2}t-{\lambda }_{1}t}}\end{array}$ (4)

As ${\lambda }_1$ and ${\lambda }_2$ is less than zero. So, ${\lambda }_1<{\lambda }_2<0$ then $0<\frac{{\lambda }_2}{{\lambda }_1}<1$ and ${\lambda }_2-{\lambda }_1=0$.

Thus the positive answer for the ‘t’ greater than 0 for the equation (4) cannot exist because the exponential (e) is increased to positive power be less than one which is not possible. Hence, the value of x is not becomes zero.

Show the graph of x versus t for the above solution as Figure (1).

(b)

To determine

Show that in case of heavy damping $\left(c>c_c\right)$ a body never passes through its position of equilibrium O started from O with an arbitrary initial velocity.

Explanation of Solution

Calculation:

Since the body is started from O with arbitrary initial velocity.

Substitute 0 for t, 0 for x and $v_0$ for v in the equation (1).

$\begin{array}{l}0=C_1{e}^{{\lambda }_1\left(0\right)}+C_2{e}^{{\lambda }_2\left(0\right)}\\ 0=C_1{e}^0+C_2{e}^0\\ 0=C_1\left(1\right)+C_2\left(1\right)\\ 0=C_1+C_2\\ C_1=-C_2\end{array}$ (5)

Substitute 0 for t, 0 for x, and $v_0$ for v in the equation (2).

$\begin{array}{l}v=\frac{dx}{dt}=C_1{\lambda }_1{e}^{{\lambda }_{1}t}+C_2{\lambda }_2{e}^{{\lambda }_{2}t}\\ v_0=C_1{\lambda }_1{e}^{{\lambda }_1\left(0\right)}+C_2{\lambda }_2{e}^{{\lambda }_2\left(0\right)}\\ v_0=C_1{\lambda }_1\left(1\right)+C_2{\lambda }_2\left(1\right)\\ v_0=C_1{\lambda }_1+C_2{\lambda }_2\end{array}$

Substitute $-C_2$ for $C_1$.

$\begin{array}{l}{C}_1{\lambda }_1+C_2{\lambda }_2=v_0\\ \left(-C_2\right){\lambda }_1+C_2{\lambda }_2=v_0\\ -C_2{\lambda }_1+C_2{\lambda }_2=v_0\\ -C_2\left(-{\lambda }_2+{\lambda }_1\right)=v_0\\ C_2\left({\lambda }_2-{\lambda }_1\right)=v_0\\ C_2=\frac{v_0}{{\lambda }_2-{\lambda }_1}\end{array}$

Substitute $\frac{v_0}{{\lambda }_2-{\lambda }_1}$ for $C_2$ in equation (5).

$C_1=-\frac{v_0}{{\lambda }_2-{\lambda }_1}$

Substitute $\frac{v_0}{{\lambda }_2-{\lambda }_1}$ for $C_2$ and $\frac{-v_0}{{\lambda }_2-{\lambda }_1}$ for $C_1$ in equation (1).

$\begin{array}{c}x=\frac{-v_0}{{\lambda }_2-{\lambda }_1}{e}^{{\lambda }_{1}t}+\frac{v_0}{{\lambda }_2-{\lambda }_1}{e}^{{\lambda }_{2}t}\\ =\frac{v_0}{{\lambda }_2-{\lambda }_1}\left[-e^{{\lambda }_{1}t}+e^{{\lambda }_{2}t}\right]\\ =\frac{v_0}{{\lambda }_2-{\lambda }_1}\left[e^{{\lambda }_{2}t}-e^{{\lambda }_{1}t}\right]\end{array}$

Apply boundary condition.

For $x=0$ when $t>0$.

$\begin{array}{l}0=\frac{v_0}{{\lambda }_2-{\lambda }_1}\left[e^{{\lambda }_{2}t}-e^{{\lambda }_{1}t}\right]\\ e^{{\lambda }_{2}t}-e^{{\lambda }_{1}t}=0\\ e^{{\lambda }_{2}t}=e^{{\lambda }_{1}t}\end{array}$

For $\left(c>c_c\right)$, the wavelengths ${\lambda }_1\ne {\lambda }_2$, thus no solution can exists for ‘t’ and x is not becomes zero when ‘t’ is greater than zero.

Show the graph of x versus t for the above solution as Figure (2).