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A particle executes SHM of amplitude A. The distance from the mean position when it's kinetic energy becomes equal to its potential energy is
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{2}} \mathrm{~A}$
Let the distance from the mean position is $\mathrm{X}$.
Given $\mathrm{KE}=\mathrm{PE}$
So, $\frac{1}{2} M \omega^2\left(A^2-x^2\right)=\frac{1}{2} M \omega^2 x^2$
$A^2-x^2=x^2 \Rightarrow A^2=2 \times 2$
$\therefore \mathrm{x}= \pm \frac{\mathrm{A}}{\sqrt{2}}$
Given $\mathrm{KE}=\mathrm{PE}$
So, $\frac{1}{2} M \omega^2\left(A^2-x^2\right)=\frac{1}{2} M \omega^2 x^2$
$A^2-x^2=x^2 \Rightarrow A^2=2 \times 2$
$\therefore \mathrm{x}= \pm \frac{\mathrm{A}}{\sqrt{2}}$
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