Question:
In an experiment for determining the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure.
What is the value of g at the place ?
In an experiment for determining the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure.
What is the value of g at the place ?
Show Hint
\(T^2 = \frac{4\pi^2}{g}L\), so g = \(4\pi^2\) divided by the slope of the \(T^2\)-L graph.
Updated On: Nov 11, 2026
- \(9.81\ \mathrm{m/s^2}\)
- \(9.87\ \mathrm{m/s^2}\)
- \(9.91\ \mathrm{m/s^2}\)
- \(10.0\ \mathrm{m/s^2}\)
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The Correct Option is B
Solution and Explanation
\(T = 2\pi\sqrt{L/g}\) so \(T^2 = \frac{4\pi^2}{g}L\); the slope of the \(T^2\) vs L line is \(4\pi^2/g\).
From the graph: L = 0.5 m gives \(T^2 = 2.0\ \mathrm{s^2}\); L = 2.0 m gives \(T^2 = 8.0\ \mathrm{s^2}\). Slope \(= \frac{8.0 - 2.0}{2.0 - 0.5} = 4\ \mathrm{s^2/m}\).
\(g = \frac{4\pi^2}{4} = \pi^2 = 9.87\ \mathrm{m/s^2}\).
Answer: (2) \(9.87\ \mathrm{m/s^2}\)
From the graph: L = 0.5 m gives \(T^2 = 2.0\ \mathrm{s^2}\); L = 2.0 m gives \(T^2 = 8.0\ \mathrm{s^2}\). Slope \(= \frac{8.0 - 2.0}{2.0 - 0.5} = 4\ \mathrm{s^2/m}\).
\(g = \frac{4\pi^2}{4} = \pi^2 = 9.87\ \mathrm{m/s^2}\).
Answer: (2) \(9.87\ \mathrm{m/s^2}\)
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