Chapter 19.4, Problem 19.110P

(a)

To determine

Find the amplitude of the motion $\left(x_m\right)$ of the bob.

Answer to Problem 19.110P

The amplitude of the motion $\left(x_m\right)$ of the bob is $\underset{\_}{1.034\text{in}\text{.}}$.

Explanation of Solution

Given information:

The weight of the bob $\left(W_B\right)$ is $2.75\text{lb}$.

The weight of the collar $\left(W_C\right)$ is $3\text{lb}$.

The length of the simple pendulum (l) is $24\text{in}\text{.}$ or $2\text{ft}$.

The amplitude of the collar $\left(x_C\right)$ is ${\delta }_m\sin {\omega }_{f}t$.

The magnitude of static deflection $\left({\delta }_m\right)$ $0.4\text{in}\text{.}$ or $0.03333\text{ft}$.

The frequency $\left(f_f\right)$ is $0.5\text{Hz}$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the mass of the bob $\left(m_B\right)$ using the formula:

$\begin{array}{c}{W}_B=mg\\ m_B=\frac{W_B}{g}\end{array}$

Substitute $2.75\text{lb}$ for $W_B$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{2.75\text{lb}}{32.2\text{ft}/{\text{s}}^{\text{2}}}\\ =0.0854\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Calculate the mass of the collar $\left(m_C\right)$ using the formula:

$m_C=\frac{W_C}{g}$

Substitute $2.75\text{lb}$ for $W_C$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_C=\frac{2.75\text{lb}}{32.2\text{ft}/{\text{s}}^{\text{2}}}\\ =0.0854\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Calculate the frequency of the period $\left({\omega }_f\right)$ force using the formula:

$f_f=\frac{{\omega }_f}{2\pi }$

Substitute $0.5\text{Hz}$ for $f_f$.

$\begin{array}{l}0.5\text{Hz}=\frac{{\omega }_f}{2\pi }\\ {\omega }_f=2\pi \left(0.5\right)\\ {\omega }_f=\pi \text{rad}/\text{s}\end{array}$

Show the system at before and after moving the collar horizontally as in Figure 1.

Refer Figure (1), $\theta$ is the angular displacement of the pendulum and mg is the gravitational force due to the mass of the bob.

Before giving horizontal movement, the force mg is equal to the tension in the pendulum.

The expression for the force balance equation for initial condition as follows:

$T=mg$

Calculate the value of $\sin \theta$ by referring Figure (1).

$\sin \theta =\frac{x-x_c}{l}$

The expression for the force balance equation in x-direction in the displaced condition as follows:

${\displaystyle \sum F_x}=m{a}_x$

Here, ${\displaystyle \sum F_x}$ is the sum of forces in x-direction and $a_x$ is the acceleration in the x-direction.

The only force in the x-direction is the tension component $T\sin \theta$ and it is opposite in direction to that of the acceleration. The acceleration in the x-direction due to the linear displacement is $\ddot{x}$.

Substitute $-T\sin \theta$ for ${\displaystyle \sum F_x}$ and $\ddot{x}$ for $a_x$.

$-T\sin \theta =m\ddot{x}$ (1)

Calculate the differential equation of motion using the formula:

Substitute mg for T and $\frac{x-x_c}{l}$ for $\sin \theta$ in equation (1).

$\begin{array}{l}-T\sin \theta =m\ddot{x}\\ -mg\left(\frac{x-x_c}{l}\right)=m\ddot{x}\\ -g\left(\frac{x-x_c}{l}\right)=\ddot{x}\\ -g\frac{x}{l}+g\frac{x_c}{l}=\ddot{x}\\ \ddot{x}+\frac{g}{l}x=\frac{g}{l}{x}_c\end{array}$

Substitute ${\delta }_m\sin {\omega }_{f}t$ for $x_c$.

$\ddot{x}+\frac{g}{l}x=\frac{g}{l}{\delta }_m\sin {\omega }_{f}t$ (2)

Here, ${\delta }_m$ is the amplitude of deflection and t is the time.

Compare the differential equation (2) with the general differential equation of motion for forced vibration $\left(\ddot{x}+{\omega }_n^{2}x={\omega }_n^2{\delta }_m\sin {\omega }_{f}t\right)$.

Calculate the natural circular frequency of vibration $\left({\omega }_n\right)$ as follows:

${\omega }_n^2=\frac{g}{l}$

Substitute $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g and $2\text{ft}$ for l.

$\begin{array}{c}{\omega }_n^2=\frac{32.2\text{ft}/{\text{s}}^{\text{2}}}{2\text{ft}}\\ =16.1\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Calculate the amplitude of the forced vibration $\left(x_m\right)$ using the formula:

$x_m=\frac{{\delta }_m}{1-\frac{{\omega }_f^2}{{\omega }_n^2}}$

Substitute $0.03333\text{ft}$ for ${\delta }_m$, $\pi$ for ${\omega }_f$, and $16.1\text{rad}/{\text{s}}^{\text{2}}$ for ${\omega }_n^2$.

$\begin{array}{c}{x}_m=\frac{0.03333\text{ft}}{1-\frac{{\pi }^2}{16.1\text{rad}/{\text{s}}^{\text{2}}}}\\ =\frac{0.03333}{0.38698}\\ =0.08613\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\\ =1.034\text{in}\text{.}\end{array}$

Therefore, the amplitude of the motion $\left(x_m\right)$ of the bob is $\underset{\_}{1.034\text{in}\text{.}}$.

(b)

To determine

Find the force that must be applied to collar C (F) to maintain the motion.

Answer to Problem 19.110P

The force that must be applied to collar C (F) to maintain the motion is $\underset{\_}{-0.103\sin \pi t\text{lb}}$.

Explanation of Solution

Given information:

The weight of the bob $\left(W_B\right)$ is $2.75\text{lb}$.

The weight of the collar $\left(W_C\right)$ is $3\text{lb}$.

The length of the simple pendulum (l) is $24\text{in}\text{.}$ or $2\text{ft}$.

The amplitude of the collar $\left(x_C\right)$ is ${\delta }_m\sin {\omega }_{f}t$.

The magnitude of static deflection $\left({\delta }_m\right)$ $0.4\text{in}\text{.}$ or $0.03333\text{ft}$.

The frequency $\left(f_f\right)$ is $0.5\text{Hz}$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Consider the collar as a free body and show the free body diagram equation as in Figure (2).

Refer Figure (2), F is the force applied to the collar to maintain the motion is F and ${\ddot{x}}_c$ is the acceleration of the collar.

The expression for the force balance equation from Figure 2 as follows:

$F+T\sin \theta =m_c{\ddot{x}}_c$

Substitute mg for T and $\frac{x-x_c}{l}$ for $\sin \theta$.

$\begin{array}{l}F+\left(mg\right)\left(\frac{x-x_c}{l}\right)=m_c{\ddot{x}}_c\\ F=-m\frac{g}{l}\left(x-x_c\right)+m_c{\ddot{x}}_c\end{array}$

Substitute ${\omega }_n^2$ for $\frac{g}{l}$.

$\begin{array}{c}F=-m\frac{g}{l}\left(x-x_c\right)+m_c{\ddot{x}}_c\\ =-m\left({\omega }_n^2\right)\left(x-x_c\right)+m_c{\ddot{x}}_c\\ =-m{\omega }_n^2\left(x-x_c\right)+m_c{\ddot{x}}_c\end{array}$ (5)

The expression for the relation for $\left(x_c\right)$ as follows:

$x_c={\delta }_m\sin {\omega }_{f}t$

Differentiate the relation for $x_c$ with respect to t.

$\begin{array}{c}{x}_c={\delta }_m\sin {\omega }_{f}t\\ {\dot{x}}_c={\delta }_m\left({\omega }_f\cos {\omega }_{f}t\right)\\ {\ddot{x}}_c={\delta }_m\left({\left({\omega }_f\right)}^2\left(-\sin {\omega }_{f}t\right)\right)\\ {\ddot{x}}_c=-{\delta }_m{\omega }_f^2\sin {\omega }_{f}t\end{array}$

Substitute $x_m\sin {\omega }_{f}t$ for x, ${\delta }_m\sin {\omega }_{f}t$ for $x_c$, and $-{\delta }_m{\omega }_f^2\sin {\omega }_{f}t$ for ${\ddot{x}}_c$ in equation (5).

$\begin{array}{c}F=-m{\omega }_n^2\left(x-x_c\right)+m_c{\ddot{x}}_c\\ =-m{\omega }_n^2\left(\left(x_m\sin {\omega }_{f}t\right)-\left({\delta }_m\sin {\omega }_{f}t\right)\right)+m_c\left(-{\delta }_m{\omega }_f^2\sin {\omega }_{f}t\right)\\ =-m{\omega }_n^2{x}_m\sin {\omega }_{f}t+m{\omega }_n^2{\delta }_m\sin {\omega }_{f}t-m_c{\delta }_m{\omega }_f^2\sin {\omega }_{f}t\\ =\left(-m{\omega }_n^2{x}_m+m{\omega }_n^2{\delta }_m-m_c{\delta }_m{\omega }_f^2\right)\sin {\omega }_{f}t\end{array}$ (6)

Substitute $0.0854\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for m, $16.1\text{rad}/{\text{s}}^{\text{2}}$ for ${\omega }_n^2$, 0.08613 ft for $x_m$, 0.03333 ft for ${\delta }_m$, $0.0854\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_c$, and $\pi \text{rad}/\text{s}$ for ${\omega }_f$ in equation (6).

$\begin{array}{c}F=\left(-m{\omega }_n^2{x}_m+m{\omega }_n^2{\delta }_m-m_c{\delta }_m{\omega }_f^2\right)\sin {\omega }_{f}t\\ =\left\{\begin{array}{l}-\left(0.0854\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\left(16.1\text{rad}/{\text{s}}^{\text{2}}\right)\left(0.08613\text{ft}\right)+\\ \left(0.0854\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\left(16.1\text{rad}/{\text{s}}^{\text{2}}\right)\left(0.03333\text{ft}\right)-\\ \left(0.0854\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\left(0.03333\text{ft}\right){\left(\pi \text{rad}/\text{s}\right)}^2\end{array}\right\}\sin \left(\pi \text{rad}/\text{s}\right)t\\ =\left(-0.1184+0.0458-0.03\right)\sin \pi t\\ =-0.103\sin \pi t\text{lb}\end{array}$

Therefore, the force that must be applied to collar C (F) to maintain the motion is $\underset{\_}{-0.103\sin \pi t\text{lb}}$.