A particle performs S.H.M. Its potential ener | vedclass

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Question

(A) The potential energy of a particle in $S.H.M.$ is given by $U = \frac{1}{2} k x^{2}$,where $k$ is the force constant.Given $U_{1} = \frac{1}{2} k

Vedclass · Accepted Answer

$\sqrt{U} = \sqrt{U_{1}} + \sqrt{U_{2}}$

Vedclass · Answer

(A) The potential energy of a particle in $S.H.M.$ is given by $U = \frac{1}{2} k x^{2}$,where $k$ is the force constant.Given $U_{1} = \frac{1}{2} k x_{1}^{2}$,we have $\sqrt{U_{1}} = \sqrt{\frac{1}{2} k} |x_{1}|$.Given $U_{2} = \frac{1}{2} k x_{2}^{2}$,we have $\sqrt{U_{2}} = \sqrt{\frac{1}{2} k} |x_{2}|$.At displacement $x = x_{1} + x_{2}$,the potential energy $U$ is $U = \frac{1}{2} k (x_{1} + x_{2})^{2}$.Taking the square root,$\sqrt{U} = \sqrt{\frac{1}{2} k} |x_{1} + x_{2}|$.Using the triangle inequality or assuming $x_{1}, x_{2}$ have the same sign,$\sqrt{U} = \sqrt{\frac{1}{2} k} |x_{1}| + \sqrt{\frac{1}{2} k} |x_{2}| = \sqrt{U_{1}} + \sqrt{U_{2}}$.Thus,$\sqrt{U} = \sqrt{U_{1}} + \sqrt{U_{2}}$.

$A$ particle performs $S.H.M.$ Its potential energies are $U_{1}$ and $U_{2}$ at displacements $x_{1}$ and $x_{2}$ respectively. At displacement $(x_{1} + x_{2})$,its potential energy $U$ is:

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