A particle executing linear SHM has a maximum velocity of `40 cm s^(-1)` and a maximum acceleration of `50 cm s^(-2)`. Find is amplitude and the period of oscillaton. Maximum velocity,

Explanation

Maximum velocity
`v_("max")= omega A=40 cm^(-1)`
Maximum acceleration,
`a_("max")= omega^(2)A =50 cm s^(-1)`
`:. (a_("max"))/(v_("max"))=(omega^(2)A)/(omegaA)=(50)/(40)`
(or) `omega=(5)/(4) "rad"s^(-1)`
Amplitude, `A=(v_("max"))/(omega)=(40xx4)/(5)=32 cm`
Period of oscillation,
`T=(2pi)/(omega)=(2xx3.142xx4)/(5)=5.032 s`