A particle perfoms a S.H.M. of amplitude A and period T. It the particle is half way between the mean position and the extreme position, then its speed at that point will be

Options

A. `(pi Asqrt(3))/(T)`
B. `(3pi^(2)A)/(T)`
C. `(pi A)/(T)`
D. `(pi Asqrt(3))/(2T)`

Answer

A

Explanation

`v=omegasqrt(A^(2)-(A^(2))/(4)) therefore v= (2pi)/(T)sqrt((3A^(2))/(4))=(piAsqrt(3))/(T)`