Potential energy of a particle in SHM along x - axis is gives by
`U = 10 + (x - 2)^(2)`
Here, `U` is in joule and `x` in metre. Total mechanical energy of the particle is `26 J`. Mass of the particle is `2kg`. Find
(a) angular frequency of SHM,
(b) potential energy and kinetic energy at mean position and extreme position,
(c ) amplitude of oscillation,
(d) x - coordinates between which particle oscillates.

Answer

A, B, C, D

Explanation

`U_(0)` = minimum potential energy at mean position

`= 10 J`
At extreme position `U` = Total mechanical energy
`= 26 J = 10 + (x - 2)^(2)`
`:. (x - 2) = +- 4`
Hence, `x = 6 m` and `x = - 2m` are the extreme position
(a) `K_(max) = E - U_(0) = 16J`
`:. (1)/(2) m omega^(2) A^(2) = 16`
or `(1)/(2) xx 2xx omega ^(2) xx (4)^(2) = 16`
or `omega = 1 rad//s`
(b) At mean position `E = 26 J`
`U = U_(0) = 10J`
`:. K = 16 J`
At extreme position
`K = 0`
`:. U = E = 26 J`.