A particle performing linear S.H.M. has a maximum displacement of 0.1 m. Its acceleration at a distance of 0.03 m from its position is 0.12 `m//s^2`. What is its velocity at a distance of 0.06 m from its mean position ?

Explanation

`v=omegasqrt(a^2-x^2) = sqrt(ac c^n/x)xsqrt(a^2-x^2)`
`sqrt0.12/0.13xxsqrt((0.1)^2-(0.06)^2)`
`2sqrt(0.01-0.0036)`
`2sqrt0.0064=2xx0.08=0.16m/s.`