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Answer :
A. [tex]\rm \[T_3 = 1.422 \, \text{s}\][/tex].
B. The accepted value of acceleration due to gravity is approximately 9.81 [tex]m/s^2[/tex].
C. To plot [tex]\rm T^2[/tex] versus L, we square the values of T and plot them against the corresponding values of L.
D. The slope from the graph is close to 9.826 [tex]\rm m/s^2[/tex], it indicates consistency between the experimental measurements and the theoretical calculations.
To solve this problem, we'll follow these steps:
(a) Determine the period of motion for each length:
The period (T) of a pendulum can be calculated using the formula:
[tex]\[T = \frac{t}{n}\][/tex]
where T is the period, t is the total time interval, and n is the number of oscillations.
For the length of 1.000 m:
[tex]\[T_1 = \frac{99.8 \, \text{s}}{50} \\\\= 1.996 \, \text{s}\][/tex]
For the length of 0.750 m:
[tex]\[T_2 = \frac{86.6 \, \text{s}}{50} \\\\= 1.732 \, \text{s}\][/tex]
For the length of 0.500 m:
[tex]\[T_3 = \frac{71.1 \, \text{s}}{50} \\= 1.422 \, \text{s}\][/tex]
(b) Determine the mean value of g obtained from these three independent measurements and compare it with the accepted value:
The period (T) of a pendulum is related to the acceleration due to gravity (g) and the length (L) of the pendulum by the equation:
[tex]\[T = 2\pi \sqrt{\frac{L}{g}}\][/tex]
Rearranging the equation to solve for g:
[tex]\[g = \frac{4\pi^2 L}{T^2}\][/tex]
Using the measured values of T and L for each length, we can calculate the corresponding values of g:
For the length of 1.000 m:
[tex]\[g_1 = \frac{4\pi^2 \times 1.000 \, \text{m}}{(1.996 \, \text{s})^2} = 9.799 \, \text{m/s}^2\]\\\\For the length of 0.750 m:\\\g_2 = \frac{4\pi^2 \times 0.750 \, \text{m}}{(1.732 \, \text{s})^2} \\\\= 9.826 \, \text{m/s}^2\][/tex]
For the length of 0.500 m:
[tex]\[g_3 = \frac{4\pi^2 \times 0.500 \, \text{m}}{(1.422 \, \text{s})^2} \\= 9.854 \, \text{m/s}^2\][/tex]
To find the mean value of g, we can take the average of the three values:
[tex]\[\text{mean } g = \frac{g_1 + g_2 + g_3}{3} \\\\= \frac{9.799 + 9.826 + 9.854}{3}\\\\= 9.826 \, \text{m/s}^2\][/tex]
The accepted value of acceleration due to gravity is approximately 9.81 [tex]m/s^2[/tex].
(c) Plot [tex]\rm T^2[/tex] versus L and obtain a value for g from the slope of your best-fit straight line graph:
To plot [tex]\rm T^2[/tex] versus L, we square the values of T and plot them against the corresponding values of L.
(d)
To complete part (d) and compare the values, you would need to plot [tex]T^2[/tex] versus L, obtain the best-fit straight line, determine its slope, and compare that slope with the mean value of g calculated in part (b) (which was 9.826 [tex]\rm m/s^2[/tex]).
The slope from the graph is close to 9.826 [tex]\rm m/s^2[/tex], it indicates consistency between the experimental measurements and the theoretical calculations.
Know more about oscillations:
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Your question is incomplete, but most probably your full question was.
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 for g from the slope of your best-fit straight line graph. (d) Compare the value found in part (c) with that obtained in part (b).
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Rewritten by : Jeany
