Question

write an expression for time period of a simple pendulumn​

Answer 1

Answer:

I don't know ...... .......

Answer 2

Answer:

$f = - mg \sin(x ) \\ according \: to \: ii \: nd \: law \\ f = ma \\ ma = - mg \sin(x) \\ a = - g \sin(x) \\ tita \: is \: very \: small \\tita = \frac{x}{l} \\ time \: period \\ t = 2\pi \sqrt{ \frac{x}{ |a| } } \\ t = 2\pi \sqrt{ \frac{x}{ | - g \frac{x}{l} | } } \\ t = 2\pi \sqrt{ \frac{xl}{gx} } \\ t = 2\pi \sqrt{ \frac{l}{g} }$

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