Two springs with spring constants m ` K _(1) = 1500 N//m`and m `K_(2) = 3000 N//m` are stretched by the same force. The ratio of potential energy stored in spring will be

Explanation

The work done in pulling the string is stored as potential energy in the spring
`U=(1)/(2)ks^(2)`……….(i)
where, k is spring constant and x is distance through which it is pulled
Also in SHM, force `prop` displacement
`implies` F=kx
Putting, `x=(F)/(k)` in Eq. (i) we get
`U=(1)/(2)k((F)/(k))^(2)=(F^(2))/(2k)`.........(ii)
`:.(U_(1))/(U_(2))=(k_(2))/(k_(1))=(3000)/(1500)(2)/(1)`
or `U_(1):U_(2)=2:1`