The displacement of a particle executiing a linear S.H.M. is given by
x=0.5 "sin" [44 t+ 0.25] metre.
Then the period and the maximum velocity of the partile are given by

Options

A. `(1)/(3)s, 44 m//s`
B. `(1)/(5)s, 66 m//s`
C. `(1)/(7)s, 22 m//s`
D. `(1)/(2)s ,33 m//s`

Answer

C

Explanation

`x=0.5 "sin" [44 t+0.25]m therefore omega t =44t therefore omega =44`
`therefore T=(2pi)/(omega)=(2xx 22)/(7xx44)=(1)/(7)s`
`therefore V_("max")= omega A=44 xx 0.5=22 m//s`