A particle starts S.H.M. from mean position along straight line and comes to rest momentraily at x = 1 m and t = 1 s. If the motion is simple harmonic, then the maximum acceleration will be `(pi^(2)=10)`

Options

A. `1 m//s^(2)`
B. `4 m//s^(2)`
C. `2.5 m//s^(2)`
D. `7 m//s^(2)`

Answer

C

Explanation

T = 4 s and A = 1m.
`therefore` The maximum acceleration `a_(m) = omega^(2)A`
`=(4pi^(2))/(T^(2))xxA=(4pi^(2))/(16)xx1=(10)/(4)=2.5`