A particle executes S.H.M with a period of 10 seconds. Find the time in which its potential energy will be half of its total energy.

Explanation

Given : ` T = 10s `
Since, ` ( 1 )/( 2 ) ` T.E. = P.E.
` therefore ( 1 )/ ( 2 ) xx ( 1 ) /( 2 ) ka ^ 2 = ( 1 ) / ( 2 ) kx ^ 2 `
` ( 1 ) /( 4 ) ka ^ 2 = ( 1 ) /( 2 ) kx ^ 2 `
` rArr x = pm ( a ) /( sqrt 2) `
Also, ` x = a sin omega t `
` a sin omega t = ( a ) /( sqrt2 ) `
` sin omega t = ( 1 ) /( sqrt 2 ) `
` sin ((2pi ) /( T) ) t = ( 1 ) / ( sqrt 2 ) `
` sin ((2pi ) /( 10 ) ) t = ( 1 )/( sqrt 2 )" "[ because T = 10 `sec]
` ( pi ) /( 5 ) t = sin ^( - 1 ) (( 1 ) /( sqrt 2 ) ) `
` (( pi ) /( 5)) t = ( pi ) /( 4 ) `
` t = ( pi ) /( 4 ) xx ( 5 )/( pi ) `
` t = 1.25 ` sec