Chapter 15, Problem 15.44P

A small object is attached to the end of a string to form a simple pendulum. The period of its harmonic motion is measured for small angular displacements and three lengths. For lengths of 1.000 m, 0.750 m, and 0.500 m, total time intervals for 50 oscillations of 99.8 s, 86.6 s, and 71.1 s are measured with a stopwatch, (a) Determine the period of motion for each length, (b) Determine the mean value of g obtained from these three independent measurements and compare it with the accepted value, (c) Plot T² versus L and obtain a value tor g from the slope of your best-fit straight-line graph. (d) Compare the value found in pan (c) with that obtained in pan (b).

To determine

The period of motion for each length of simple pendulum.

Answer to Problem 15.44P

The period of motion for the length $1.00\text{m}$ of simple pendulum is $1.996\text{s}$, the period of motion for the length $0.75\text{m}$ of simple pendulum is $1.732\text{s}$ and the period of motion for the length $0.50\text{m}$ of simple pendulum is $1.422\text{s}$.

Explanation of Solution

Given info: The lengths of the simple pendulum are $1.00\text{m}$, $0.75\text{m}$, $0.50\text{m}$ and the total time intervals for $50\text{oscillations}$ of $99.8\text{s}$, $86.6\text{s}$, $71.1\text{s}$ are measured with a stopwatch.

For the $1.00\text{m}$ length of the pendulum, the total time interval of $50\text{oscillations}$ is $99.8\text{s}$, so the period of per oscillation is the ration of total time interval and the number of oscillations,

$\begin{array}{c}{T}_1=\frac{99.8\text{s}}{50}\\ =1.996\text{s}\end{array}$

Here,

$T_1$ is the period of per oscillation of $1.00\text{m}$ length of the pendulum.

Thus, the period for the $1.00\text{m}$ length of the pendulum is $1.996\text{s}$.

For the $0.75\text{m}$ length of the pendulum, the total time interval of $50\text{oscillations}$ is $86.6\text{s}$, so the period of per oscillation is the ration of total time interval and the number of oscillations,

$\begin{array}{c}{T}_2=\frac{86.6\text{s}}{50}\\ =1.732\text{s}\end{array}$

Here,

$T_2$ is the period of per oscillation of $0.75\text{m}$ length of the pendulum.

Thus, the period for the $0.75\text{m}$ length of the pendulum is $1.732\text{s}$.

For the $0.50\text{m}$ length of the pendulum, the total time interval of $50\text{oscillations}$ is $71.1\text{s}$, so the period of per oscillation is the ration of total time interval and the number of oscillations,

$\begin{array}{c}{T}_3=\frac{71.1\text{s}}{50}\\ =1.422\text{s}\end{array}$

Here,

$T_3$ is the period of per oscillation of $0.50\text{m}$ length of the pendulum.

Thus, the period for the $0.50\text{m}$ length of the pendulum is $1.422\text{s}$.

Conclusion:

Therefore, the period of motion for the length $1.00\text{m}$ of simple pendulum is $1.996\text{s}$, the period of motion for the length $0.75\text{m}$ of simple pendulum is $1.732\text{s}$ and the period of motion for the length $0.50\text{m}$ of simple pendulum is $1.422\text{s}$.

To determine

The mean value of $g$ obtained from these three independent measurements and compare it with the accepted values.

Answer to Problem 15.44P

The mean value of $g$ for the length $1.00\text{m}$ of simple pendulum is $9.9{\text{m/s}}^2$, the mean value of $g$ for the length $0.75\text{m}$ of simple pendulum is $9.87{\text{m/s}}^2$ and the mean value of $g$ for the length $0.50\text{m}$ of simple pendulum is $9.76{\text{m/s}}^2$.

Explanation of Solution

Given info: The lengths of the simple pendulum are $1.00\text{m}$, $0.75\text{m}$, $0.50\text{m}$ and the total time intervals for $50\text{oscillations}$ of $99.8\text{s}$, $86.6\text{s}$, $71.1\text{s}$ are measured with a stopwatch.

The period of the oscillation of the pendulum is,

$T=2\pi \sqrt{\frac{L}{g}}$

Here,

$T$ is the period of the oscillations.

$L$ is the length of the pendulum.

$g$ is the acceleration due to gravity.

Take square on the both sides and calculate the $g$ from the above equation,

$\begin{array}{c}{T}^2={\left(2\pi \sqrt{\frac{L}{g}}\right)}^2\\ T^2=4{\pi }^2\left(\frac{L}{g}\right)\\ g=4{\pi }^2\left(\frac{L}{T^2}\right)\end{array}$ (1)

Substitute $1.00\text{m}$ for $L$ and $1.996\text{s}$ for $T$ in above equation.

$\begin{array}{c}g=4{\pi }^2\left(\frac{1.00\text{m}}{{\left(1.996\text{s}\right)}^2}\right)\\ =4{\pi }^2\left(0.25{\text{m/s}}^2\right)\\ =9.9{\text{m/s}}^2\end{array}$

Thus, the mean value of $g$ for the length $1.00\text{m}$ of simple pendulum is $9.9{\text{m/s}}^2$.

Substitute $0.75\text{m}$ for $L$ and $1.732\text{s}$ for $T$ in above equation.

$\begin{array}{c}g=4{\pi }^2\left(\frac{0.75\text{m}}{{\left(1.732\text{s}\right)}^2}\right)\\ =4{\pi }^2\left(0.433{\text{m/s}}^2\right)\\ =9.87{\text{m/s}}^2\end{array}$

Thus, the mean value of $g$ for the length $0.75\text{m}$ of simple pendulum is $9.87{\text{m/s}}^2$.

Substitute $0.50\text{m}$ for $L$ and $1.422\text{s}$ for $T$ in above equation.

$\begin{array}{c}g=4{\pi }^2\left(\frac{0.50\text{m}}{{\left(1.422\text{s}\right)}^2}\right)\\ =4{\pi }^2\left(0.351{\text{m/s}}^2\right)\\ =9.76{\text{m/s}}^2\end{array}$

Thus, the mean value of $g$ for the length $0.50\text{m}$ of simple pendulum is $9.76{\text{m/s}}^2$.

Conclusion:

Therefore, the mean value of $g$ for the length $1.00\text{m}$ of simple pendulum is $9.9{\text{m/s}}^2$, the mean value of $g$ for the length $0.75\text{m}$ of simple pendulum is $9.87{\text{m/s}}^2$ and the mean value of $g$ for the length $0.50\text{m}$ of simple pendulum is $9.76{\text{m/s}}^2$.

To determine

To draw: The graph of $T^2$ versus $L$ and find the value of $g$ from the slope of best-fit straight line graph.

Answer to Problem 15.44P

The graph of $T^2$ versus $L$ is shown in below figure,

Figure (1)

The value of the $g$ from the slope of the above graph is $10.1{\text{m/s}}^2$.

Explanation of Solution

Given info: The lengths of the simple pendulum are $1.00\text{m}$, $0.75\text{m}$, $0.50\text{m}$ and the total time intervals for $50\text{oscillations}$ of $99.8\text{s}$, $86.6\text{s}$, $71.1\text{s}$ are measured with a stopwatch.

In the part (a), the periods of different given length of pendulum are calculated. Make a table of square of periods $T^2$ for given three lengths of pendulum.

$L$	$T^2$
$1.00\text{m}$	${\left(1.996\text{s}\right)}^2=3.96{\text{s}}^2$
$0.75\text{m}$	${\left(1.732\text{s}\right)}^2=2.99{\text{s}}^2$
$0.50\text{m}$	${\left(1.422\text{s}\right)}^2=2.022{\text{s}}^2$

The above table gives ordered pairs.

Take the ordered pairs from above given table and join them by a straight line and plot the graph of $T^2$ versus $L$,

Figure (1)

The Figure (1) shows the graph of $T^2$ versus $L$ as straight line.

From the above graph, the slope of the line is,

$m=\frac{y_2-y_1}{x_2-x_1}$

Here,

$m$ is the slope of the line.

$x_1,x_2$ are the two points at $x$-axis.

$y_1,y_2$ are the two points at $y$-axis.

Substitute $1.00\text{m}$ for $x_1$, $0.75\text{m}$ for $x_2$, $3.96{\text{s}}^2$ for $y_1$ and $2.99{\text{s}}^2$ for $y_2$ in above equation.

$\begin{array}{c}m=\frac{2.99{\text{s}}^2-3.96{\text{s}}^2}{1.00\text{m}-0.75\text{m}}\\ =\frac{-0.97{\text{s}}^2}{-0.25\text{m}}\\ =3.88{\text{s}}^2\text{/m}\end{array}$

From the equation of the period of the pendulum,

$T^2=\frac{4{\pi }^{2}L}{g}$

The slope of $T^2$ versus $L$ is $\frac{4{\pi }^2}{g}$.

So, $\frac{4{\pi }^2}{g}$ is equal to the $3.88{\text{s}}^2\text{/m}$ slope of the graph.

$\begin{array}{l}\frac{4{\pi }^2}{g}=3.88{\text{s}}^2\text{/m}\\ g=\frac{4{\pi }^2}{3.88{\text{s}}^2\text{/m}}\\ =10.1{\text{m/s}}^2\end{array}$

To determine

The comparison of values of $g$ from part (c) and from part (b).

Answer to Problem 15.44P

The value of $g$ from part (c) is $10.1{\text{m/s}}^2$ and the average value of $g$ from part (b) of different length of the pendulum is $9.85{\text{m/s}}^2$ and there is no difference between them.

Explanation of Solution

Given info: The lengths of the simple pendulum are $1.00\text{m}$, $0.75\text{m}$, $0.50\text{m}$ and the total time intervals for $50\text{oscillations}$ of $99.8\text{s}$, $86.6\text{s}$, $71.1\text{s}$ are measured with a stopwatch.

From part (c) the value of $g$ is $10.1{\text{m/s}}^2$ and the average value of $g$ from part (b) of different lengths of the pendulum is $9.85{\text{m/s}}^2$.

Both values of $g$ have not major difference, both are approximately equal.

Conclusion:

Therefore, the value of $g$ from part (c) is $10.1{\text{m/s}}^2$ and the average value of $g$ from part (b) of different length of the pendulum is $9.85{\text{m/s}}^2$ and there is no difference between them.