Chapter 15, Problem 44P

(a)

To determine

The period of motion of pendulum for each lengths.

Answer to Problem 44P

The period of motion for length $1.000\text{m}$ is $\underset{\_}{2.00\text{s}}$, $0.750\text{m}$ is $\underset{\_}{1.73\text{s}}$, and for length $0.500\text{m}$ is $\underset{\_}{1.42\text{s}}$.

Explanation of Solution

Write the expression for the time period of oscillation of a simple pendulum.

    $T=\frac{t}{n}$                                                                                                                         (I)

Here, $T$ is the time period of oscillation of simple pendulum, $t$ is the measured time of oscillation, and $n$ is the number of oscillations.

Conclusion:

Consider simple pendulum of length $1.000\text{m}$.

Substitute $99.8\text{s}$ for $t$, and $50$ for $n$ in equation (I) to find $T$.

    $\begin{array}{c}T=\frac{99.8\text{s}}{50}\\ =1.999\text{s}\\ \approx \text{2}\text{.00}\text{s}\end{array}$

Consider simple pendulum of length $0.750\text{m}$.

Substitute $86.6\text{s}$ for $t$, and $50$ for $n$ in equation (I) to find $T$.

    $\begin{array}{c}T=\frac{86.6\text{s}}{50}\\ =1.73\text{s}\end{array}$

Consider simple pendulum of length $0.500\text{m}$.

Substitute $71.1\text{s}$ for $t$, and $50$ for $n$ in equation (I) to find $T$.

    $\begin{array}{c}T=\frac{71.1\text{s}}{50}\\ =1.42\text{s}\end{array}$

Tabulated form of the length of pendulum and the corresponding time periods is shown in Table 1 below.

Therefore, the period of motion for length $1.000\text{m}$ is $\underset{\_}{2.00\text{s}}$, $0.750\text{m}$ is $\underset{\_}{1.73\text{s}}$, and for length $0.500\text{m}$ is $\underset{\_}{1.42\text{s}}$.

(b)

To determine

The average value of $g$ for each pendulums and compare it with the accepted value.

Answer to Problem 44P

The average value of $g$ obtained is $\underset{\_}{9.85{\text{m/s}}^2}$, and it is in agreement with the accepted value within $\underset{\_}{0.5\%}$.

Explanation of Solution

Write the expression for the time period of the simple pendulum.

    $T_i=2\pi \sqrt{\frac{l_i}{g_i}}$                                                                                                              (II)

Here, $T_i$ is the time period, $l_i$ is the length of the pendulum, and $g_i$ is the acceleration due to gravity.

Square expression (II) and rearrange to find $g$.

    $g_i=\frac{4{\pi }^2{L}_i}{T_i^2}$                                                                                                              (III)

Write the expression to find the average acceleration due to gravity.

    $\begin{array}{c}{g}_{av}=\frac{{\displaystyle \sum _{i=1}^n{g}_i}}{n}\\ =\frac{g_1+g_2+g_3}{3}\end{array}$                                                                                                   (IV)

Here, $g_{av}$ is the average acceleration due to gravity, and $n$ is the number of values measured.

Write the expression to find the degree of closeness of the average value to the accepted value of acceleration due to gravity.

    $g\%=\frac{g_{av}}{g}\times 100\%$                                                                                                    (V)

Here, $g\%$ is the closeness of $g_{av}$ to accepted value, and $g$ is the accepted value.

Conclusion:

Consider simple pendulum of length $1.000\text{m}$. Its time period is $2.00\text{s}$. So put $i=1$.

Substitute $2.00\text{s}$ for $T_1$, and $1.000\text{m}$ for $L_1$ in equation (III) to get $g_1$.

    $\begin{array}{c}{g}_1=\frac{4{\pi }^2\left(1.000\text{m}\right)}{{\left(2.00\text{s}\right)}^2}\\ =9.87{\text{m/s}}^2\end{array}$

Consider simple pendulum of length $0.750\text{m}$. Its time period is $1.73\text{s}$. Put $i=2$.

Substitute $1.73\text{s}$ for $T_2$, and $0.750\text{m}$ for $L_2$ in equation (III) to get $g_2$.

    $\begin{array}{c}{g}_2=\frac{4{\pi }^2\left(0.750\text{m}\right)}{{\left(1.73\text{s}\right)}^2}\\ =9.89{\text{m/s}}^2\end{array}$

Consider simple pendulum of length $0.500\text{m}$. Its time period is $1.42\text{s}$. Put $i=3$.

Substitute $1.42\text{s}$ for $T_3$, and $0.500\text{m}$ for $L_3$ in equation (III) to get $g_3$.

    $\begin{array}{c}{g}_3=\frac{4{\pi }^2\left(0.500\text{m}\right)}{{\left(1.42\text{s}\right)}^2}\\ =9.79{\text{m/s}}^2\end{array}$

The tabulate form of values of time period, length and acceleration due to gravity is given in Table 2 below.

Substitute $9.87{\text{m/s}}^2$ for $g_1$, $9.89{\text{m/s}}^2$ for $g_2$, and $9.79{\text{m/s}}^2$ for $g_3$ in equation (IV) to find $g_{av}$.

    $\begin{array}{c}{g}_{av}=\frac{9.87{\text{m/s}}^2+9.89{\text{m/s}}^2+9.79{\text{m/s}}^2}{3}\\ =9.85{\text{m/s}}^2\end{array}$

Substitute $9.85{\text{m/s}}^2$ for $g_{av}$, and $9.80{\text{m/s}}^2$ for $g$ in equation (V) to find $g\%$.

    $\begin{array}{c}g\%=\frac{9.85{\text{m/s}}^2}{9.80{\text{m/s}}^2}\times 100\%\\ =0.50\%\end{array}$

Therefore, the average value of $g$ obtained is $\underset{\_}{9.85{\text{m/s}}^2}$, and it is in agreement with the accepted value within $\underset{\_}{0.5\%}$.

(c)

To determine

Plot $T^2$ versus $L$ graph and obtain the value of $g$ from the slope of the graph.

Answer to Problem 44P

Figure 1 gives the plot between $T^2$ and $L$ and the value of acceleration due to gravity from the slope of graph is $\underset{\_}{9.94{\text{m/s}}^2}$.

Explanation of Solution

From Table 2 obtain the values for length of pendulums and the corresponding time period of oscillations. Find the value of $T^2$. Plot $T^2$ against $L$.

Plot $T^2$ along the $Y$ axis and $L$ along $X$ axis of the graph. Take the values of $L$ from $0\text{m}\text{to 1}\text{.000}\text{m}$ and the values of $T^2$ from $0\text{s to 4}\text{.00}\text{s}$. The resultant graph is a straight line.

Write the expression for $T^2$.

    $T^2=\left(\frac{4{\pi }^2}{g}\right)L$                                                                                                         (VI)

From equation (VI), slope of the graph is $\frac{4{\pi }^2}{g}$.

    $m=\frac{4{\pi }^2}{g}$                                                                                                               (VII)

Here, $m$ is the slope of the graph.

Rearrange expression (VII) to find $g$.

    $g=\frac{4{\pi }^2}{m}$                                                                                                               (VIII)

Conclusion:

The slope of graph is got as $3.97{\text{s}}^2\text{/m}$. Substitute $3.97{\text{s}}^2\text{/m}$ for $m$ in equation (VIII) to find $g$.

    $\begin{array}{c}g=\frac{4{\pi }^2}{3.97{\text{s}}^2\text{/m}}\\ =9.94{\text{m/s}}^2\end{array}$

Therefore, Figure 1 gives the plot between $T^2$ and $L$ and the value of acceleration due to gravity from the slope of graph is $\underset{\_}{9.94{\text{m/s}}^2}$.

(d)

To determine

The level of closeness of $g$ from graph to the accepted value of $g$.

Answer to Problem 44P

The value of $g$ from graph is $1.5\%$ close to the accepted value of $g$.

Explanation of Solution

Write the expression to find the level of closeness of acceleration due to gravity from graph and accepted value of $g$.

    $g\%=\frac{g_g}{g}\times 100\%$                                                                                                     (IX)

Here, $g_g$ is the value of acceleration due to gravity from the graph.

Conclusion:

Substitute $9.94{\text{m/s}}^2$ for $g$, and $9.80{\text{m/s}}^2$ for $g$ in equation (IX) to find $g\%$.

    $\begin{array}{c}g\%=\frac{9.94{\text{m/s}}^2}{9.80{\text{m/s}}^2}\times 100\%\\ =1.5\%\end{array}$

Therefore, the value of $g$ from graph is $1.5\%$ close to the accepted value of $g$.