Chapter 19.1, Problem 19.16P

(a)

To determine

The time $\left(\tau \right)$ required for the bob to return to point A.

Answer to Problem 19.16P

The time $\left(\tau \right)$ required for the bob to return to point A is $\underset{\_}{1.876\sec }$.

Explanation of Solution

Given Information:

The cord length $\left(l_{AB}\right)$ of the bob is 1.2 m.

At rest the angle $\left({\theta }_A\right)$ is $5°$.

The distance (d) is 0.6 m.

Assuming the value of acceleration due to gravity (g) is $9.81{\text{m/s}}^{\text{2}}$.

Calculation:

Calculate natural circular frequency for path ${\left({\omega }_n\right)}_{AB}$ between the points A and B using the relation:

${\left({\omega }_n\right)}_{AB}=\sqrt{\frac{g}{l_{AB}}}$

Substitute $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for g and 1.2 m for $l_{AB}$.

$\begin{array}{c}{\left({\omega }_n\right)}_{AB}=\sqrt{\frac{\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}}{1.2\text{m}}}\\ =\sqrt{8.175}\\ =2.8592\text{rad}/\text{s}\end{array}$

The time period of oscillations corresponding to the natural circular frequency ${\left({\omega }_n\right)}_{AB}$ is given as $2\pi /{\left({\omega }_n\right)}_{AB}$. This time period is the time taken to complete one full oscillation, but the time period required is only for the path A to B and then from B to A. Half the time period of $2\pi /{\left({\omega }_n\right)}_{AB}$ is the time period for the path between the points A and B.

Calculate the time period $\left({\tau }_{AB}\right)$ of oscillations for the path between the points A and B using the relation:

${\tau }_{AB}=\frac{1}{2}\left(\frac{2\pi }{{\left({\omega }_n\right)}_{AB}}\right)$

Substitute $2.8592\text{rad}/\text{s}$ for ${\left({\omega }_n\right)}_{AB}$.

$\begin{array}{c}{\tau }_{AB}=\frac{1}{2}\left(\frac{2\pi }{2.8592\text{rad}/\text{s}}\right)\\ =1.09876\text{s}\end{array}$

Calculate the length $\left(l_{BC}\right)$ of the cord for the path between the points B and C using the relation:

$l_{BC}=\left(1.2-d\right)\text{m}$

Substitute 0.6 m for d.

$\begin{array}{c}{l}_{BC}=\left(1.2-0.6\right)\text{m}\\ =\text{0}\text{.6}\text{m}\end{array}$

Calculate the natural circular frequency ${\left({\omega }_n\right)}_{BC}$ for path between the points B and C using the relation:

${\left({\omega }_n\right)}_{BC}=\sqrt{\frac{g}{l_{BC}}}$

Substitute $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for g and 0.6 m for $l_{BC}$.

$\begin{array}{c}{\left({\omega }_n\right)}_{BC}=\sqrt{\frac{\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}}{0.6\text{m}}}\\ =\sqrt{16.35}\\ =4.0435\text{rad}/\text{s}\end{array}$

The time period of oscillations corresponding to the natural circular frequency ${\left({\omega }_n\right)}_{BC}$ is given as $2\pi /{\left({\omega }_n\right)}_{BC}$. This time period is the time taken to complete one full oscillation, but the time period required is only for the path B to C and then from C to B. Half the time period of $2\pi /{\left({\omega }_n\right)}_{BC}$ is the time period for the path between the points B and C.

Calculate the time period $\left({\tau }_{BC}\right)$ of oscillations for the path between the points B and C using the relation:

${\tau }_{BC}=\frac{1}{2}\left(\frac{2\pi }{{\left({\omega }_n\right)}_{BC}}\right)$

Substitute $4.0435\text{rad}/\text{s}$ for ${\left({\omega }_n\right)}_{BC}$.

$\begin{array}{c}{\tau }_{BC}=\frac{1}{2}\left(\frac{2\pi }{4.0435\text{rad}/\text{s}}\right)\\ =0.77695\text{s}\end{array}$

Calculate the time period to return to A using the relation:

$\tau ={\tau }_{AB}+{\tau }_{BC}$

Substitute 1.09876 s for ${\tau }_{AB}$ and 0.77695 s for ${\tau }_{BC}$.

$\begin{array}{c}\tau =\left(1.09876\text{s}\right)+\left(0.77695\text{s}\right)\\ =1.876\text{s}\end{array}$

Therefore, the time $\left({\tau }_n\right)$ required for the bob to return to point A is $\underset{\_}{1.876\sec }$.

(b)

To determine

The amplitude $\left({\theta }_C\right)$.

Answer to Problem 19.16P

The amplitude $\left({\theta }_C\right)$ is $\underset{\_}{7.07°}$.

Explanation of Solution

Given Information:

The cord length $\left(l_{AB}\right)$ of the bob is 1.2 m.

At rest the angle $\left({\theta }_A\right)$ is $5°$.

The distance (d) is 0.6 m.

Assume the acceleration due to gravity (g) is $9.81{\text{m/s}}^{\text{2}}$.

Calculation:

For the path between the points A and B:

Consider point A:

The displacement at A $\left({\theta }_A\right)$ is the maximum displacement $\left({\theta }_{m1}\right)$. Hence, the ${\theta }_A$ is equal to ${\theta }_{m1}$.

Consider point B:

Express the derivative of the displacement at B $\left({\dot{\theta }}_B\right)$:

${\dot{\theta }}_B={\left({\omega }_n\right)}_{AB}{\theta }_{m1}$

Substitute ${\theta }_A$ for ${\theta }_{m1}$.

${\dot{\theta }}_B={\left({\omega }_n\right)}_{AB}{\theta }_A$ (1)

Express the velocity $\left(v_B\right)$ at B:

$v_B=l_{AB}{\dot{\theta }}_B$

Substitute Equation (1) for the value of ${\dot{\theta }}_B$.

$\begin{array}{c}{v}_B=l_{AB}{\dot{\theta }}_B\\ =l_{AB}\left({\left({\omega }_n\right)}_{AB}{\theta }_A\right)\\ =l_{AB}{\left({\omega }_n\right)}_{AB}{\theta }_A\end{array}$ (2)

For the path between the points B and C:

Consider point C:

The displacement at C $\left({\theta }_C\right)$ is the maximum displacement $\left({\theta }_{m2}\right)$. Thus, the ${\theta }_C$ is equal to ${\theta }_{m2}$.

Consider point B:

Express the derivative of the displacement at B:

${\dot{\theta }}_B={\left({\omega }_n\right)}_{BC}{\theta }_B$

Substitute ${\theta }_C$ for ${\theta }_B$.

${\dot{\theta }}_B={\left({\omega }_n\right)}_{BC}{\theta }_C$ (3)

Express the velocity at B:

$v_B=l_{BC}{\dot{\theta }}_B$

Substitute Equation (3) for the value of ${\dot{\theta }}_B$.

$\begin{array}{c}{v}_B=l_{BC}{\dot{\theta }}_B\\ =l_{BC}\left({\left({\omega }_n\right)}_{BC}{\theta }_C\right)\\ =l_{BC}{\left({\omega }_n\right)}_{BC}{\theta }_C\end{array}$ (4)

Equate Equations (2) and (4).

$\begin{array}{l}{l}_{BC}{\left({\omega }_n\right)}_{BC}{\theta }_C=l_{AB}{\left({\omega }_n\right)}_{AB}{\theta }_A\\ {\theta }_C=\frac{l_{AB}{\left({\omega }_n\right)}_{AB}}{l_{BC}{\left({\omega }_n\right)}_{BC}}{\theta }_A\end{array}$ (5)

Calculate the amplitude $\left({\theta }_C\right)$:

Substitute 1.2 m for $l_{AB}$, 0.6 m for $l_{BC}$, $2.8592\text{rad}/\text{s}$ for ${\left({\omega }_n\right)}_{AB}$, $4.0435\text{rad}/\text{s}$ for ${\left({\omega }_n\right)}_{BC}$, and $5°$ for ${\theta }_A$ in Equation (5).

$\begin{array}{c}{\theta }_C=\frac{\left(1.2\text{m}\right)\left(2.8592\text{rad}/\text{s}\right)}{\left(0.6\text{m}\right)\left(4.0435\text{rad}/\text{s}\right)}\left(5°\right)\\ =\left(1.4142\right)\left(5°\right)\\ =7.07°\end{array}$

Therefore, the amplitude $\left({\theta }_C\right)$ is $\underset{\_}{7.07°}$.