The time period of a particle in simple harmonic motion is T. Assume potential energy at mean position to be zero. After a time of `(T)/(6)` it passes its mean position its,

Options

A. velocity will be half its maximum velocity
B. displacement will be half its amplitude
C. acceleration will be nearly `86%` of its maximum
D. `KE=PE`

Answer

A, C

Explanation

`y=asinomegat=asin((2pit)/(T))`
`v=(dy)/(dt)=omegaacos((2pit)/(T))`
At `t=(T)/(6)`,`v=omegaacos(((2pi)/(T)(T)/(6)))=(1)/(2)omegaa`
or, `v=((v_(max))/(2))`
`y=asin((2pi)/(T))xx(T)/(6)=asin(pi)/(3)`
It is not half of a.
acceleration `=(d^2y)/(dt^2)=(dv)/(dt)=omega^2asin((2pit)/(T))`
`=omega^2asin((pi)/(3))=0.86(AC)_(max)`
At this instant,
`KE=(1)/(2)mv^2=(1)/(2)m((v_(max))/(2))^2=((KE)_(max))/(4)=(1)/(4)(TE)`
`PE=TE-KE=(3)/(4)(TE)`
i.e., `KEnePE`