In an experiment for determining the gravitat | vedclass

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(B) From the given graph,we can observe the relationship between $T^2$ and $L$. Taking a point from the line,for $L = 1.0 \ m$,the corresponding value

Vedclass · Accepted Answer

$9.87$

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(B) From the given graph,we can observe the relationship between $T^2$ and $L$. Taking a point from the line,for $L = 1.0 \ m$,the corresponding value is $T^2 = 4.0 \ s^2$. The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$. Squaring both sides,we get $T^2 = 4\pi^2 \frac{L}{g}$. Rearranging for $g$,we get $g = \frac{4\pi^2 L}{T^2}$. Substituting the values from the graph: $g = \frac{4 	imes (3.14)^2 	imes 1.0}{4.0} = \pi^2 \approx 9.87 \ m/s^2$. Thus,the value of $g$ is $9.87 \ m/s^2$.

In an experiment for determining the gravitational acceleration $g$ of a place with the help of a simple pendulum,the measured time period square $(T^2)$ is plotted against the string length $(L)$ of the pendulum as shown in the figure. What is the value of $g$ at the place in $m/s^2$?

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