The total energy of a particle executing $S.H.M.$ is $80 \,J$. What is the potential energy when the particle is at a distance of $\frac{3}{4}$ of amplitude from the mean position..... $J$
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(d) $\frac{U}{E} = \frac{{\frac{1}{2}m{\omega ^2}{y^2}}}{{\frac{1}{2}m{\omega ^2}{a^2}}} = \frac{{{y^2}}}{{{a^2}}}$
==> $\frac{U}{{80}} = \frac{{{{\left( {\frac{3}{4}a} \right)}^2}}}{{{a^2}}} = \frac{9}{{16}}$
==> $U = 45\,J$
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