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CGP EDU Academic Team
Published on: September 12, 2026
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 3/4 of amplitude from the mean position
Text Solution
Verified by ExpertsThe correct answer is:
D
$\frac{U}{E} = \frac{\frac{1}{2}m\omega^{2}y^{2}}{\frac{1}{2}m\omega^{2}a^{2}} = \frac{y^{2}}{a^{2}} \Rightarrow \frac{U}{80} = \frac{\left(\frac{3}{4}a\right)^{2}}{a^{2}} = \frac{9}{16} \Rightarrow U = 45J$
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