A particle performs S.H.M. Its potential energies are 'U $_{1}$ ' and ${ }^{\prime} \mathrm{U}_{2}$ ' at…

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
  1. $\sqrt{\mathrm{U}}=\sqrt{\mathrm{U}_{1}}+\sqrt{\mathrm{U}_{2}}$
  2. $\sqrt{\mathrm{U}}=\left(\sqrt{\mathrm{U}_{1}}+\sqrt{\mathrm{U}_{2}}\right)^{2}$
  3. $\sqrt{\mathrm{U}}=\sqrt{\mathrm{U}_{1}}-\sqrt{\mathrm{U}_{2}}$
  4. $\sqrt{\mathrm{U}}=\left(\sqrt{\mathrm{U}_{1}}-\sqrt{\mathrm{U}_{2}}\right)^{2}$

Solution

$\sqrt{E_{1}}+\sqrt{E_{2}}=\sqrt{E}$ .

Asked in: MHT CET 2020 (15 Oct Shift 1)

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