question_answer

Two springs of spring constants 1000 N/m and 4000 N/m respectively are stretched with the same force, their potential energies will be in ratio:

A)  1 : 2         

B)  4 : 1

C)  1 : 4         

D)  16 : 1

Correct Answer: B

Solution :

Key Idea: Spring mass system oscillates in SHM. The force required to stretch the spring is \[F=ky\] where y is displacement. Given, \[{{F}_{1}}={{F}_{2}}\] \[\therefore \] \[{{k}_{1}}{{y}_{1}}={{k}_{2}}{{y}_{2}}\] \[\Rightarrow \] \[\frac{{{y}_{1}}}{{{y}_{2}}}=\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{4000}{1000}=\frac{4}{1}\] The work done in stretching the spring is stored as potential energy. \[U=\frac{1}{2}k{{y}^{2}}\] \[\therefore \] \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{{{k}_{1}}y_{1}^{2}}{{{k}_{2}}y_{2}^{2}}=\frac{{{k}_{1}}}{{{k}_{2}}}{{\left( \frac{{{y}_{1}}}{{{y}_{2}}} \right)}^{2}}\] \[=\frac{{{k}_{1}}}{{{k}_{2}}}{{\left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)}^{2}}=\frac{1}{4}{{\left( \frac{4}{1} \right)}^{2}}=\frac{4}{1}\]