Question

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

Answer 1

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`