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                                                             ?(i) \[U=\frac{1}{2}k{{x}^{2}}\]                                          where k is spring constant and \[x\] is distance through which it is pulled. Also in SHM Force \[\propto \]displacement \[F=kx\]                                               ?(ii) where, k is spring constant. Putting \[x=\frac{F}{k}\]in Eq. (i), we get                 \[U=\frac{1}{2}k{{\left( \frac{F}{R} \right)}^{2}}=\frac{{{F}^{2}}}{2k}\]                 \[\therefore \]  \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{3000}{1500}=\frac{2}{1}\] \[\therefore \]  \[{{U}_{1}}:{{U}_{2}}=2:1\]