A)
B)
C)
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Correct Answer: C
Solution :
Potential energy is given by \[U=\frac{1}{2}k{{x}^{2}}\] The corresponding graph is shown in figure. At equilibrium position (x = 0), potential energy is minimum. At extreme positions \[{{x}_{1}}\] and \[{{x}_{2}}\], its potential energies arc \[{{U}_{1}}=\frac{1}{2}\,kx_{1}^{2}\,and\,\,{{U}_{2}}=\frac{1}{2}\,kx_{2}^{2}\] Note: In the above graph, the dotted line (curve) is shown for kinetic energy. This graph shows that kinetic energy is maximum at mean position and zero at extreme positions \[{{x}_{1}}\] and \[{{x}_{2}}\].You need to login to perform this action.
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