The equation of motion of a particle executing SHM is `x=asin(omegat+pi/6)` with time period T. Find the time interval at which the velocity is being half of its maximum value.

Explanation

Displacement of the particle executing simple harmonic monic motion,
`x=asin(omegat+pi/6)`
`therefore"Velocity",v=(dx)/(dt)=aomegacos(omegat+pi/6)`
`therefore` Maximum velocity, `v_(max)=aomega`
Here, `v=v_(max)/2`
or, `aomegacos(omegat+pi/6)=(aomega)/2`
or, `cos(omegat+pi/6)=1/2="cos"pi/3`
`therefore" "omegat+pi/6=pi/3or,(2pi)/T*t=pi/6" "[becauseomega=(2pi)/T]`
`therefore" "t/T=1/12`
So, after `1/12` of time period, the velocity of the particle is being half of its maximum value.