question_answer

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

A)  1 : 2                                      

B)  1 : 4

C)  4 : 1                                      

D)  2 : 1

Correct Answer: A

Solution :

                 When a body oscillates in simple 'harmonic motion, it is acted upon by a restoring force which tends to bring it in the equilibrium position. Due to this force there is potential energy in the body. Restoring force acting on body is F = mass \[\times \] acceleration \[=m{{\omega }^{2}}x\] where,\[\omega \]is angular  frequency and \[x\] is displacement. The work done in displacing the body appears as potential energy U. \[U=W=\int_{0}^{x}{m{{\omega }^{2}}xdx=m{{\omega }^{2}}\frac{{{x}^{2}}}{2}}\] Here, mo\[m{{\omega }^{2}}=k=\] spring constant. \[\therefore \]    \[{{U}_{1}}=\frac{1}{2}{{k}_{1}}{{x}^{2}},{{U}_{2}}=\frac{1}{2}{{k}_{2}}{{x}^{2}}\] \[\therefore \]      \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{1500}{3000}=\frac{1}{2}\] \[\therefore \]    \[{{U}_{2}}:{{U}_{2}}=1:2\]