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A particle performs S.H.M. from the mean position. Its amplitude is 'A' and total
energy is 'E'. At a perticular instant its kinetic energy is $\frac{3 \mathrm{E}}{4}$. The displacement of the
particle at that instant is
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energy is 'E'. At a perticular instant its kinetic energy is $\frac{3 \mathrm{E}}{4}$. The displacement of the
particle at that instant is
Solution:
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Verified Answer
The correct answer is:
$\frac{A}{2}$
Total energy $=\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$
When kinetic energy is $\frac{3 \mathrm{E}}{4}$, its potential energy is $\left(\mathrm{E}-\frac{3 \mathrm{E}}{4}\right)=\frac{\mathrm{E}}{4}$
Dividing Eq.(1) by Eq.(2), $\quad 4=\frac{A^{2}}{x^{2}} \quad \therefore x=\frac{A}{2}$
When kinetic energy is $\frac{3 \mathrm{E}}{4}$, its potential energy is $\left(\mathrm{E}-\frac{3 \mathrm{E}}{4}\right)=\frac{\mathrm{E}}{4}$
Dividing Eq.(1) by Eq.(2), $\quad 4=\frac{A^{2}}{x^{2}} \quad \therefore x=\frac{A}{2}$
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