A particle executing S.H.M. has a maximum speed of 30`cm//s` and a maximum acceleration of `60cm//s^(2)`. The period of oscillation is

Options

A. `pi sec`
B. `(pi)/(2)sec`
C. `2pisec`
D. `(pi)/(t)sec`

Answer

A

Explanation

Maximum speed, `v_(max) = omegaA …(i)`
Maximum acceleration, `a_(max) = omega^(2)A ….(ii)`
Divide eqn. (ii) by eqn. (i), we get …(iii)
`(a_(max))/(v_(max)) = (omega^(2)A)/(omegaA) = omega :. (a_(max))/(v_(max)) = (2T)/(T) T = 2pi ((v_(max))/(a_(max)))`
Here, `v_(max) =30 cms^(-1), a_(max) = 60 cms^(-2)`
`:. T = 2pi ((30cms^(-1))/(60cms^(-2))) = pis`