A particle executes a S.H.M. of amplitude A and maximum velocity `V_(m)` then its speed at displacement `A//2` is

Options

A. `(V_("max"))/(2)`
B. `(V_("max"))/(4)`
C. `V_("max")`
D. `0.866 V_("max")`

Answer

D

Explanation

`v=omegasqrt(A^(2)-x^(2))= omega sqrt(A^(2)-(A^(2))/(4))`
`=omega sqrt((3A^(2))/(4))=(sqrt(3))/(2)omega A`
Maximum velocity `= omega A`
`therefore v=(sqrt(3))/(2)(v_("max"))=0.866(v_("max"))`