Question

what is the time period of simple pendulum?

Answer 1

the period of swing of a simple gravity pendulum depends on its lenght ,the local strength of gravity ,and to a small extent on the maximum angle that the pendulum swings away from vertical ,called the amplitude.It is independent of the mass of the bob.

Answer 2

Answer:

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$\underline {\bold{Time\:Period\:of\:Simple\:Pendulum}}\\\\\sf{Consider}\:simple\:bob\:attached\:to\:a\:string\:of\:length\:\bold{L}.\\\\Forces\:acting\:are:-weight(mg)\:and\:Tension(T).\\\\Restoring\:force\:,F=-mgsinx\\\\If\:y\:is\:small\:sinx=x.\\\implies\:F=-mgx\\\\Angle,\:x=\dfrac{Arc\:length}{Radius}=\dfrac{BA}{L}=\dfrac{Y}{L}\\\\\therefore\:F=mg*\dfrac{Y}{L}\\\\By\:Newton's\:law\:F=ma\\\implies\:ma=-mg*\dfrac{Y}{L}\\\\a=-\dfrac{gY}{L}\implies\:a\:\propto\:-Y,is \:an\:SHM$

$\sf{a=-{\omega}^{2}Y}\\\\\implies\:-g\bigg(\dfrac{Y}{L}\bigg)=-{\omega}^{2}Y\\\\\implies\:\dfrac{g}{L}={\omega}^{2}\\\\\implies\:\omega=\sqrt{\dfrac{g}{L}}\:\implies\:\dfrac{2\pi}{T}=\sqrt{\dfrac{g}{L}}\\\\\\\therefore\:\bold{Time\:Period(T)=2\pi\sqrt{\dfrac{L}{g}}}$