Chapter 19.4, Problem 19.110P

To determine

(a)

The amplitude of the motion of bob

Answer to Problem 19.110P

The amplitude of motion of bob is $1.034\text{in}\text{.}$.

Explanation of Solution

Given:

Mass of bob is $2.75\text{ lb}$.

Length of pendulum is $24\text{ in}\text{.}$.

Mass of collar is $3\text{ lb}$.

The relation of displacement is $x_c={\delta }_m\sin \left({\omega }_{f}t\right)$.

The maximum amplitude of displacement is $0.4\text{ in}\text{.}$.

Frequency is $0.5\text{ Hz}$.

Concept used:

Draw the diagram for system as shown below:

As the bob has a small displacement, from the above figure,

$\sin \theta =\frac{\left(x-x_c\right)}{l}$

Here, $x$ is the final displacement of bob after the angular displacement, $x_c$ is the displacement and $l$ is the length of pendulum.

Write the expression for the force balance in initial condition.

$T=mg$

Here, $T$ is the tension in string, $m$ is the mass of bob and $g$ is the acceleration due to gravity.

Write the expression for Newton’s Law of motion in x-direction.

$\sum F_x=m{a}_x$

Substitute $-T\sin \theta$ for $\sum F_x$ and $\ddot{x}$ for $a_x$ in the above expression.

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

Here, $\theta$ is the small angular displacement and $\ddot{x}$ is then acceleration in x-direction.

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

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

Substitute ${\delta }_m\sin \left({\omega }_{f}t\right)$ for $x_c$ in the above expression and rearrange.

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

Write the general differential equation of motion.

$\ddot{x}+{\omega }_n^{2}x=F$ ...... (3)

Here, ${\omega }_n$ is the natural circular frequency.

Compare equation (3) and (2).

${\omega }_n^2=\frac{g}{l}$ ...... (4)

Write the expression for the amplitude of forces vibration.

$x_m=\frac{{\delta }_m}{1-\left(\frac{{\omega }_f^2}{{\omega }_n^2}\right)}$ ...... (5)

Here, $x_m$ is the amplitude of vibration, ${\delta }_m$ is the maximum amplitude of displacement and ${\omega }_f$ is the forced frequency of vibration.

Calculation:

Substitute $32.2{\text{ ft/s}}^2$ for $g$ and $24\text{ in}\text{.}$ for $l$ in equation (4).

$\begin{array}{c}{\omega }_n^2=\frac{32.2}{24\text{ in}\text{.}\left(\frac{1\text{ ft}}{12\text{ in}\text{.}}\right)}\\ =16.1\text{ rad/s}\end{array}$

Substitute $16.1\text{ rad/s}$ for ${\omega }_n^2$ and $0.4\text{ in}\text{.}$ for ${\delta }_m$ and $\pi$ for ${\omega }_f$ in equation (5).

$\begin{array}{c}{x}_m=\frac{\left(0.4\text{ in}\text{.}\right)}{1-\left(\frac{{\pi }^2}{16.1}\right)}\\ =1.034\text{in}\text{.}\end{array}$

The amplitude of motion of bob is $1.034\text{in}\text{.}$.

Conclusion:

Thus, the amplitude of motion of bob is $1.034\text{in}\text{.}$.

To determine

(b)

The periodic force that should be applied to the collar to maintain periodic motion.

Answer to Problem 19.110P

The periodic force required to maintain the motion of collar is $-0.103\text{sin}\left(\pi t\right)\text{ lb}$.

Explanation of Solution

Concept used:

Draw the FBD of collar as shown below:

In the above Figure $F$ is the force applied to maintain the motion, ${\ddot{x}}_c$ is the acceleration of collar and $m_c$ is the mass of collar.

Write the expression for Newton’s Law of motion in x-direction.

$\sum F_x=m_c{a}_x$

Substitute $F+T\sin \theta$ for $\sum F_x$ and ${\ddot{x}}_c$ for $a_x$ in the above expression.

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

Substitute $mg$ for $T$ and $\frac{\left(x-x_c\right)}{l}$ for $\sin \theta$ in the above expression.

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

Substitute ${\omega }_n^2$ for $\frac{g}{l}$ in above equation and simplify for $F$.

$F=-m{\omega }_n^2\left(x-x_c\right)+m_c{\ddot{x}}_c$ ...... (6)

Write the relation for $x_c$.

$x_c={\delta }_m\sin \left({\omega }_{f}t\right)$

Differentiate the above expression with respect to $t$.

${\dot{x}}_c={\delta }_m{\omega }_f\left(\cos {\omega }_{f}t\right)$

Differentiate the above expression with respect to $t$.

${\ddot{x}}_c=-{\delta }_m{\omega }_f^2\sin \left({\omega }_{f}t\right)$

Substitute $x_m\sin \left({\omega }_{f}t\right)$ for $x$, ${\delta }_m\sin \left({\omega }_{f}t\right)$ for $x_c$ and $-{\delta }_m{\omega }_f^2\sin \left({\omega }_{f}t\right)$ for ${\ddot{x}}_c$ in equation (6).

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

Simplify the above expression for $F$.

`

$F=\left(-m{\omega }_n^2{x}_m+m{\omega }_n^2{\delta }_m-m_c{\delta }_m{\omega }_f^2\right)\sin \left({\omega }_{f}t\right)$ ...... (7)

Calculation:

Substitute $0.0854{\text{lb-s}}^{\text{2}}\text{/ft}$ for $m$, $16.1\text{ rad/s}$ for ${\omega }_n^2$, $0.08613\text{ ft}$ for $x_m$, $0.03333\text{ ft}$ for ${\delta }_m$, $0.09316{\text{ lb-s}}^{\text{2}}\text{/ft}$ for $m_c$ and $\pi \text{ rad/s}$ for ${\omega }_f$ in equation (7).

$\begin{array}{c}F=\left\{\begin{array}{l}-\left(0.0854\right)\left(16.1\right)\left(0.08613\right)+\left(0.0854\right)\left(16.1\right)\left(0.03333\right)\\ -\left(0.09316\right)\left(0.03333\right){\pi }^2\end{array}\right\}\sin \left(\pi t\right)\\ =\left(-0.1184+0.0458-0.030\right)\sin \left(\pi t\right)\\ =-0.103\sin \left(\pi t\right)\text{ lb}\end{array}$

The periodic force required to maintain the motion of collar is $-0.103\text{sin}\left(\pi t\right)\text{ lb}$.

Conclusion:

Thus, the periodic force required to maintain the motion of collar is $-0.103\text{sin}\left(\pi t\right)\text{ lb}$.