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what is the period, in seconds, of a simple pendulum of length 5 meters?use the gravitational constant g=9.8 m/s^2 and round your answer to two decimal places
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\[T=2\Pi \sqrt{l/g}\] where T is time period, l is length of pendulum g is gravitational constant.. so by calculation, T=4.48 sec
\[\large T=2\pi \sqrt{\frac{l}{g}}\]\[\large T=2\pi \sqrt{\frac{5}{9.8}}\]\[\Large T=4.48s\]
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