Show that the combined spring energy and gravitational energy for a mass `m` hanging from a light spring of force constant `k` can be expressed as `U_(0) + (1)/(2) ky^(2)`, where `y` is the distance above or below the equilibrium position and `U_(0)` is constant.

Answer

A

Explanation

At equilibrium position, `kx_(0) = mg`

In displaced position, `U = (1)/(2)k (x_(0) + y)^(2) - mg y`
` = (1)/(2)kx_(0)^(2) + 1/2ky^(2) + kx_(0)y - mgy`
Substituting `kx_(0) = mg` we get,
`U = (1)/(2)kx_(0)^(2) + (1)/(2)ky^(2)`
` = (1)/(2)kx_(0)^(2) + (1)/2ky^(2) + kx_(0)y - mgy`
But `(1)/(2)kx_(0)^(2)` = constant say `U_(0)`
`:. U = U_(0) + (1)/(2) ky^(2)`.