The bob of a simple pendulum performs a S.H.M. of amplitude 2 cm. If the mass of the bob is 100 gram and its total energy is `32xx10^(-5)` J. The periodic time of S.H.M. is

Options

A. 1.2 sec
B. 2 sec
C. 2.5 sec
D. 1.571 sec

Answer

D

Explanation

`E=(1)/(2) m omega^(2)A^(2)`
`therefore 32xx10^(-5)=(1)/(2)xx100xx10^(-3)xx omega^(2)xx4xx10^(-4)`
`therefore omega^(2)=16 therefore omega =4`
`therefore (2pi)/(T)=4 therefore T=(pi)/(2)=1.571s`