Question

What is time period of simple pendulum . ?

Answer 1

Heya.....!!!

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A simple pendulum executes SHM and Time Period of Simple Pendulum is :-

$= > \: \: \: \: 2\pi \sqrt{ \frac{l}{g} }$


Here ,
• g - Acceleration due to gravity .
• l - Length of pendulum .


From this formulae we can conclude that Time period of Simple pendulum does not depend upon mass of the Bob of pendulum . Time period depends on Acceleration due to gravity and length of pendulum .


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Answer 2

Answer:

Explanation:

$\implies\:a\:\propto\:-x,is \:an\:SHM\\\\\therefore\:a=-{\omega}^{2}x\\\\\implies\:{\omega}^{2}=\dfrac{g}{L}\implies\:\omega=\sqrt{\dfrac{g}{L}}\\\\\implies\:T=2\pi*\sqrt{\dfrac{L}{g}}$$\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=-mgsiny\\\\If\:y\:is\:small\:siny=y.\\\implies\:F=-mgy\\\\Angle,\:y=\dfrac{Arc\:length}{Radius}=\dfrac{OA}{L}=\dfrac{x}{L}\\\\\therefore\:F=mg*\dfrac{x}{L}\\\\By\:Newton's\:law\:F=ma\\\implies\:ma=-mg*\dfrac{x}{L}\\\\a=\dfrac{gx}{L}\implies\:a\:\propto\:-x,is \:an\:SHM\\\\\therefore\:a=-{\omega}^{2}x$