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Q.
The total energy of a particle executing SHM is $80\,J$. What is the potential energy when the particle is ata distance of $3/4$ of amplitude from the mean position?
Oscillations
Solution:
Given, $\frac{1}{2} m \omega^{2} r^{2}=80 J ;$
$\therefore P E=\frac{1}{2} m \omega^{2} y^{2}=\frac{1}{2} m \omega^{2} \times\left(\frac{3}{4} r\right)^{2}$
$=\frac{9}{16}\left(\frac{1}{2} m \omega^{2} r^{2}\right)=\frac{9}{16} \times 80=45\, J$