A body is executing a simple harmonic motion such that its potential energy is `U_(1) `at `U_(2)at y` When the displacement is `x+y` the potential energy will be

Options

A. `U_(1) + U_(2)`
B. `sqrt(U_(1)^(2) + U_(2)^(2))`
C. `U_(1) + U_(2) + 2sqrt(U_(1) U_(2))`
D. `sqrt(U_(1) U_(2))`

Answer

C

Explanation

At displacement `x,U_(1) = (1)/(2) m omega^(2) x^(2)`
`rArr x = sqrt((2U_(1))/(m omega^(2)))`
Similarly `y = sqrt((2U_(2))/(m omega^(2)))`
Where displacement is `(x + y)` potential energy is
`U = (1)/(2) m omega^(2) (x + y)^(2) = (1)/(2) omega^(2) (x^(2) + y^(2) + 2 sqrt(xy))`
`= (1)/(2) m omega^(2)x^(2) + (1)/(2)m omega^(2) y^(2) +m omega^(2) sqrt(xy)`
`= U_(1) + U_(2)+ m omega^(2)sqrt((2U_(1))/(m omega^(2)) (2U_(2))/(m omega^(2)))`
`= U_(1) + U_(2) + 2 sqrt(U_(1)U_(2))`