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CGP EDU Academic Team
Published on: September 12, 2026
A particle starts simple harmonic motion from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is $3E/4.$ Its displacement at that instant is
Text Solution
Verified by ExpertsThe correct answer is:
B
$\frac{K}{E} = \frac{\frac{1}{2} m \omega^{2} (a^{2} - y^{2})}{\frac{1}{2} m \omega^{2} a^{2}} = \frac{a^{2} - y^{2}}{a^{2}} = 1 - \frac{y^{2}}{a^{2}}$
$So, \left(\frac{\frac{3E}{4}}{E}\right) = 1 - \frac{y^2}{a^2} \Rightarrow \frac{y^2}{a^2} = 1 - \frac{3}{4} = \frac{1}{4} \Rightarrow y = \frac{a}{2}.$
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