If stretch in a spring of force constant k is doubled, calculate
(a) ration of final to initial force in the spring.
(b) ratio of elastic energies stored in the two cases.
(c) work done in changing to the state of double stretch.

Explanation

(a) As for a spring `F=ky`
`F_(2)/F_(1)=(ky_(2))/(ky_(1))=(2y)/y=2`
(b) As for a spring `U=(1//2)ky^(2)`
`U_(2)/U_(1)=((1//2)ky_(2)^(2))/((1//2)ky_(1)^(2))=((2y)^(2))/((y)^(2))=4`
(c) As work done is stored as elastic potential energy of spring
`W=U_(2)-U_(1)=1/2k(2y)^(2)-1/2k(y)^(2)=3/2ky^(2)[=3U_(1)]`