The potential energy of a particle of mass 2 kg in SHM is `(9x^(2))`J. Here x is the displacement from mean position . If total mechanical energy of the particle is 36 J. The maximum speed of the particle is

Answer

6

Explanation

At x = 0, U = 0,Therefore, totalmechanical energy is equal to the maximum kinetic energy.
`therefore 1/2 mv_("max")^(2) = 36`
or `v_("max") = sqrt(72/m)`
`= sqrt(72/2) = 6 m//s`