Chapter 19.5, Problem 19.131P

To determine

(a)

The time interval between a maximum positive displacement and the following maximum negative displacement is $\frac{{\tau }_{\text{d}}}{2}$.

Explanation of Solution

Given information:

The system is underdamped, the time period between two successive points is ${\tau }_{\text{d}}=\frac{2\pi }{{\omega }_{\text{d}}}$.

The figure shows the underdamped system.

Figure-(1)

Write the expression amplitude of vibrations of an underdamped system..

$x=x_0{e}^{\left(\frac{-c}{2m}\right)t}\sin \left({\omega }_{\text{d}}t+\phi \right)$ ..... (I)

Here, the amplitude of an underdamped system is $x$, the maximum amplitude of an underdamped system is $x_0$, the damping coefficient is $c$, the mass of vibrating system is $m$, the damped frequency is ${\omega }_{\text{d}}$, the time period between two successive points is $t$, the phase difference is $\phi$.

Differentiate Equation (I) with respect to time.

$\begin{array}{l}\frac{dx}{dt}=\frac{d\left(x_0{e}^{\left(\frac{-c}{2m}\right)t}\sin \left({\omega }_{\text{d}}t+\phi \right)\right)}{dt}\\ V=x_0{\omega }_{\text{d}}{e}^{\left(\frac{-c}{2m}\right)t}\cos \left({\omega }_{\text{d}}t+\phi \right)+x_0\left(\frac{-c}{2m}\right)e^{\left(\frac{-c}{2m}\right)t}\sin \left({\omega }_{\text{d}}t+\phi \right)\end{array}$ ...... (II)

For maximum positive and negative displacement, $V=0$.

$\begin{array}{l}0=x_0{\omega }_{\text{d}}{e}^{\left(\frac{-c}{2m}\right)t}\cos \left({\omega }_{\text{d}}t+\phi \right)+x_0\left(\frac{-c}{2m}\right)e^{\left(\frac{-c}{2m}\right)t}\sin \left({\omega }_{\text{d}}t+\phi \right)\\ x_0{\omega }_{\text{d}}{e}^{\left(\frac{-c}{2m}\right)t}\cos \left({\omega }_{\text{d}}t+\phi \right)=x_0\left(\frac{c}{2m}\right)e^{\left(\frac{-c}{2m}\right)t}\sin \left({\omega }_{\text{d}}t+\phi \right)\\ \tan \left({\omega }_{\text{d}}t+\phi \right)=\frac{2m{\omega }_{\text{d}}}{c}\end{array}$ ...... (III)

The maximum negative displacement occurs after a phase difference of $\pi$.

$\tan \left({\omega }_{\text{d}}t+\phi -\pi \right)=\frac{2m{\omega }_{\text{d}}}{c}$ ...... (IV)

Calculation:

Substitute $t_0$ for $t$ in Equation (III).

$\begin{array}{l}\left({\omega }_{\text{d}}{t}_0\right)={\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)-\phi \\ t_0=\frac{{\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)-\phi }{{\omega }_{\text{d}}}\end{array}$ ...... (V)

Here, the time interval for maximum positive displacement is $t_0$.

Substitute $t_1$ for $t$ in Equation (IV).

$\begin{array}{l}\tan \left({\omega }_{\text{d}}{t}_1+\phi -\pi \right)=\frac{2m{\omega }_{\text{d}}}{c}\\ \left({\omega }_{\text{d}}{t}_1+\phi -\pi \right)={\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)\\ {\omega }_{\text{d}}{t}_1={\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)+\pi -\phi \\ t_1=\frac{{\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)+\pi -\phi }{{\omega }_{\text{d}}}\end{array}$ ...... (VI)

Here, the time interval for maximum negative displacement is $t_1$.

Substract Equation (V) and (VI).

$\begin{array}{l}{t}_0-t_1=\frac{{\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)-\phi }{{\omega }_{\text{d}}}-\left(\frac{{\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)+\pi -\phi }{{\omega }_{\text{d}}}\right)\\ t_0-t_1=\frac{{\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)-\phi -{\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)-\pi +\phi }{{\omega }_{\text{d}}}\\ t_0-t_1=\frac{-\pi }{{\omega }_{\text{d}}}\\ t_1-t_0=\frac{\pi }{{\omega }_{\text{d}}}\end{array}$ ...... (VII)

Substitute $\frac{2\pi }{{\tau }_{\text{d}}}$ for ${\omega }_{\text{d}}$ in Equation (VII).

$\begin{array}{l}{t}_1-t_0=\frac{\pi }{\frac{2\pi }{{\tau }_{\text{d}}}}\\ t_1-t_0=\frac{{\tau }_{\text{d}}}{2}\end{array}$

Conclusion:

The time interval between a maximum positive displacement and the following maximum negative displacement is $t_1-t_0=\frac{{\tau }_{\text{d}}}{2}$.

To determine

(b)

The time interval between two successive zero displacement is $\frac{{\tau }_{\text{d}}}{2}$

Explanation of Solution

Calculation:

Substitute $0$ for $x$ in Equation (I).

$\begin{array}{l}0=x_0{e}^{\left(\frac{-c}{2m}\right)t}\sin \left({\omega }_{\text{d}}t+\phi \right)\\ \sin \left({\omega }_{\text{d}}t+\phi \right)=0°\\ \left({\omega }_{\text{d}}t+\phi \right)={\sin }^{-1}\left(0°\right)\\ \left({\omega }_{\text{d}}t+\phi \right)=n\pi \end{array}$ ...... (VIII)

Substitute $1$ for $n$ and $t_2$ for $t$ in Equation (VIII).

$\begin{array}{l}\left({\omega }_{\text{d}}{t}_2+\phi \right)=\pi \\ {\omega }_{\text{d}}{t}_2=\pi -\phi \\ t_2=\frac{\pi -\phi }{{\omega }_{\text{d}}}\end{array}$ ...... (IX)

Here, the time interval for first zero displacement is $t_2$.

Substitute $2$ for $n$ and $t_3$ for $t$ in Equation (VIII).

$\begin{array}{l}\left({\omega }_{\text{d}}{t}_3+\phi \right)=2\pi \\ {\omega }_{\text{d}}{t}_3=2\pi -\phi \\ t_3=\frac{2\pi -\phi }{{\omega }_{\text{d}}}\end{array}$ ...... (X)

Here, the time interval for second zero displacement is $t_3$.

Substract Equation (X) from (IX).

$\begin{array}{c}{t}_3-t_2=\frac{2\pi -\phi }{{\omega }_{\text{d}}}-\left(\frac{\pi -\phi }{{\omega }_{\text{d}}}\right)\\ =\frac{2\pi -\phi -\pi +\phi }{{\omega }_{\text{d}}}\\ t_3-t_2=\frac{\pi }{{\omega }_{\text{d}}}\end{array}$ ...... (XI)

Substitute $\frac{2\pi }{{\tau }_{\text{d}}}$ for ${\omega }_{\text{d}}$ in Equation (XI).

$\begin{array}{l}{t}_1-t_0=\frac{\pi }{\frac{2\pi }{{\tau }_{\text{d}}}}\\ t_1-t_0=\frac{{\tau }_{\text{d}}}{2}\end{array}$

Conclusion:

The time interval between two successive zero displacement is $\frac{{\tau }_{\text{d}}}{2}$

To determine

(c)

The time interval between maximum positive displacement and zero displacement is greater than $\frac{{\tau }_{\text{d}}}{4}$

Explanation of Solution

Calculation:

Substract Equation (V) from (IX).

$\begin{array}{l}{t}_0-t_2=\frac{{\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)-\phi -\left(\pi -\phi \right)}{{\omega }_{\text{d}}}\\ t_0-t_2=\frac{{\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)-\pi }{{\omega }_{\text{d}}}\\ {\omega }_{\text{d}}\left(t_0-t_2\right)={\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)\\ \pi +{\omega }_{\text{d}}\left(t_0-t_2\right)={\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)-\pi \end{array}$

$\begin{array}{l}\tan \left(\pi +{\omega }_{\text{d}}\left(t_0-t_2\right)\right)=\left(\frac{2m{\omega }_{\text{d}}}{c}\right)\\ \tan \left({\omega }_{\text{d}}\left(t_0-t_2\right)\right)=\left(\frac{2m{\omega }_{\text{d}}}{c}\right)\\ {\omega }_{\text{d}}\left(t_0-t_2\right)={\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)\end{array}$

The values of ${\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)$ lies between $\left(\frac{-\pi }{2},\frac{\pi }{2}\right)$.

$\frac{-\pi }{2}<{\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)<\frac{\pi }{2}$ ...... (XII)

Substitute ${\omega }_{\text{d}}\left(t_0-t_2\right)$ for ${\tan }^{-1}\left(\frac{2m{\omega }_{\text{d}}}{c}\right)$ in Equation (XII).

$\begin{array}{l}\frac{-\pi }{2}<{\omega }_{\text{d}}\left(t_0-t_2\right)<\frac{\pi }{2}\\ \frac{-\pi }{2{\omega }_{\text{d}}}<\left(t_0-t_2\right)<\frac{\pi }{2{\omega }_{\text{d}}}\end{array}$

Neglect the time interval difference to be negative.

$\begin{array}{l}\left(t_0-t_2\right)<\frac{\pi }{2{\omega }_{\text{d}}}\\ -\left(t_2-t_0\right)<\frac{\pi }{2{\omega }_{\text{d}}}\\ \left(t_2-t_0\right)>\frac{\pi }{2{\omega }_{\text{d}}}\end{array}$ ...... (XIII)

Substitute $\frac{2\pi }{{\tau }_{\text{d}}}$ for ${\omega }_{\text{d}}$ in Equation (XIII).

$\begin{array}{l}\left(t_2-t_0\right)>\frac{\pi }{2\frac{2\pi }{{\tau }_{\text{d}}}}\\ \left(t_2-t_0\right)>\frac{{\tau }_{\text{d}}}{4}\end{array}$

Conclusion:

The time interval between maximum positive displacement and zero displacement is greater than $\frac{{\tau }_{\text{d}}}{4}$