question_answer

Two springs of spring constant 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have potential energy in ratio

A)  1 : 2        

B)  2 : 1

C)  1 : 4        

D)  4 : 1

Correct Answer: B

Solution :

Key Idea: When spring is pulled and left, it oscillation in SHM. The work done in pulling the string is stored as potential energy in the spring                 \[U=\frac{1}{2}k\,{{x}^{2}}\] ... (i) where k is spring constant and is distance through which it is  pulled. Also in SHM Force \[\propto \] displacement                 \[F=k\,x\] ... (ii) where, k is spring constant. Putting \[x=\frac{F}{k}\], in Eq. (i), we get                 \[U=\frac{1}{2}k\,\,{{\left( \frac{F}{R} \right)}^{2}}=\frac{{{F}^{2}}}{2\,k}\] \[\therefore \] \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{3000}{1500}=\frac{2}{1}\] \[\therefore \] \[{{U}_{1}}:{{U}_{2}}=2:1\]