A particle executes SHM with a time period of `4 s`. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

Explanation

Here, `a=30cm, T=4s,y=15cm. `
As `y=asinomegat=asin((2pi)/(T))t`
`15=30sin((2pi)/(4))t or sin((2pi)/(4))t=(15)/(30)=(1)/(2)=sin((pi)/(6))`
`:. (2pi)/(4)t=(pi)/(6) or t=(4)/(2xx6)=(1)/(3)s`