When a mass m is attached to the spring of force constant k, then the spring stretches by l. If the mass oscillates with amplitude l, what will be the maximum potential energy stored in the spring ?

Options

A. `(kl)/(2)`
B. 2kl
C. `(1)/(2)` mgl
D. mgl.

Answer

D

Explanation

When mass is suspended then spring stretched by l
`:." mg"="kl"` and energy stored in spring
`=(1)/(2)kl^(2)=(1)/(2)("mg")/(l)l^(2)=(1)/(2)"mgl"`.
When the spring oscillates at maximum displacement then additional potential energy stored `=(1)/(2)kl^(2)=(1)/(2)"mgl"`. Thus total potential energy `=(1)/(2)"mgl"+(1)/(2)"mgl"="mgl"`.
Correct choice is (d).