The maximum displacement of a particle executing S.H.M. is 1 cm and the maximum acceleration is `(1.57)^(2)`cm per `sec^(2)` . Then the time period is

Options

A. 0.25 cm `s^(-2)`
B. 4.0 s
C. 1.57 s
D. 3.14 s

Answer

B

Explanation

Maximum acceleration `= A omega^(2)`
`rArr ((pi)/(2))^(2) = 1 ((2pi)/(T))^(2)" "(because 1.57 = (pi)/(2))`
`rArr (pi^(2))/(4) = (4 pi^(2))/(T^(2))`
`rArr T^(2) = (4 xx 4 pi^(2))/(pi^(2))`
`T^(2) = 16`
`rArr T = 4s`