For a body executing SHM with amplitude A, time period T, maximum velocity `v_(max)` and phace constant zero, which of the following statements are correct for `0letle(T)/(4)` (y is displacement from mean position)?

Options

A. At `y=((A)/(2))`,`vgt((v_(max))/(2))`
B. For `v=((v_(max))/(2))`,`y=lt((A)/(2))`
C. For `t=((T)/(8)`,`ygt((A)/(2))`
D. for `y=((A)/(2))`,`ylt((T)/(8))`.

Answer

A, B, C, D

Explanation

For SHM
`y=Asinomegat=Asin((2pit)/(T))`
`v=omegasqrt(A^2-y^2)` , `v_(max)=omegaA`
(a) At `y=(A)/(2)`
`v=omegasqrt(A^2-(A^2)/(4))=sqrt((3omegaA)/(2))gt(omegaA)/(2)`
(b) For `v=v_((0)/(2))impliesomegasqrt(^2-y^2)=(omegaA)/(2)`
`impliesA^2-y^2=(A^2)/(4)`
for `t=(T)/(8)`
`y=Asin((2piT)/(8T))=(A)/(sqrt2)gt(A)/(sqrt2)`
(d) for `y=(A)/(2)`, `(A)/(2)=Asin((2pit)/(T))`
`implies(2pit)/(T)=(pi)/(6)=(pi)/(6)` `implies` `t=(T)/(12)lt(T)/(8)`