A particle is executing S.H.M. with amplitude 'A' and maximum velocity `V_(m)`. The displacement at which its velocity is half of the maximum velocity is

Options

A. `86.6%A`
B. `13.4%A`
C. `70.7%A`
D. `36.72%A`

Answer

A

Explanation

`v=omegasqrt(A^(2)-x^(2))`
`therefore (Aomega)/(2)=omegasqrt(A^(2)-x^(2))`
`(A^(2))/(4)-A^(2)=x^(2)therefore (3A^(2))/(4)=x^(2)`