The displacement of a particle performing a S.H.M. is given by `x=0.5 "sin" 100 pi (t + 0.05)`, where x is in metres and t is in second. Its periodic time in second is

Options

A. 0.01s
B. 0.02s
C. 0.1s
D. 0.5s

Answer

B

Explanation

`x=0.5 "sin" (100 pi t+5 pi)`
Comparing with `x=A "sin" (omega t+ alpha)`, we get, `omega =100 pi`
`therefore (2pi)/(T)=100 pi`
`therefore T=(2pi)/(100 pi)=(1)/(50)=0.02 s`