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

Options

A. `4 ms^(-1)`
B. `2 ms^(-1)`
C. `6 ms^(-1)`
D. `10 ms^(-1)`

Answer

C

Explanation

At x = 0, U = 0. Therefore total mechanical 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 ms^(-1)`