MHT CET · Physics · Oscillations
A particle performs S.H.M. Its potential energies are 'U \(_{1}\) ' and \({ }^{\prime} \mathrm{U}_{2}\) ' at displacements
\({ }^{\prime} \mathrm{x}_{1}{ }^{\prime}\) and \({ }^{\prime} \mathrm{x}_{2}\) ' respectively. At displacement \(\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)\), its potential energy ' \(\mathrm{U}^{\prime}\) is
- A \(\sqrt{\mathrm{U}}=\sqrt{\mathrm{U}_{1}}+\sqrt{\mathrm{U}_{2}}\)
- B \(\sqrt{\mathrm{U}}=\left(\sqrt{\mathrm{U}_{1}}+\sqrt{\mathrm{U}_{2}}\right)^{2}\)
- C \(\sqrt{\mathrm{U}}=\sqrt{\mathrm{U}_{1}}-\sqrt{\mathrm{U}_{2}}\)
- D \(\sqrt{\mathrm{U}}=\left(\sqrt{\mathrm{U}_{1}}-\sqrt{\mathrm{U}_{2}}\right)^{2}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\mathrm{U}}=\sqrt{\mathrm{U}_{1}}+\sqrt{\mathrm{U}_{2}}\)
Step-by-step Solution
Detailed explanation
\(\sqrt{E_{1}}+\sqrt{E_{2}}=\sqrt{E}\)
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