Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The total energy of the body executing S.H.M. is E. Then the kinetic energy when the displacement is half of the amplitude, is
$(a) \frac{E}{2} (b) \frac{E}{4} (c) \frac{3E}{4} (d) \frac{\sqrt{3}}{4} E$
Text Solution
Verified by ExpertsThe correct answer is:
C
$Total energy in SHM E = \frac{1}{2} m \omega^{2} a^{2} ; (where a = amplitude) Potential energy U = \frac{1}{2} m \omega^{2} (a^{2} - y^{2}) = E - \frac{1}{2} m \omega^{2} y^{2} When y = \frac{a}{2} \implies U = E - \frac{1}{2} m \omega^{2} \left(\frac{a^{2}}{4}\right) = E - \frac{E}{4} = \frac{3E}{4}$
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