A particle performing linear S.H.M. has a period of 6.28 seconds and a path length of 20 cm. What is the velocity when its displacement is 6 cm from mean position ?

Explanation

Given , `T = 6.28` sec, Path length, `P = 20` cm
`rArr a = P/2= 10 cm = 10 xx 10^(-12) m`
`x = 6 xx 10^(-2) m`
As we know, `omega = (2pi)/(T)`
`= (2pi)/(6.28) = 1` rad/s
`v = +- omegasqrt((100-36)xx10^(-4))`
`[ :' v = omega sqrt(a^(2) - x^(2))]`
`= 1 xx 8 xx 10^(-2)`
`:. v = 8xx 10^(-2) m//s`