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?

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(2) 9.87 m/s². Read the slope off the graph: T² rises from 2.0 s² at L=0.5 m to 8.0 s² at L=2.0 m. Slope =(8.0-2.0)/(2.0-0.5)=(6.0)/(1.5)=4.0 s² m⁻¹. From T²=(4π²)/gL, the slope is (4π²)/g. g=(4π²)/(4.0)=π²=9.87 m/s².

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?

Step-by-step solution

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