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A particle executing simple harmonic motion with amplitude $A$ has the same potential and kinetic energies at the displacement
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Verified Answer
The correct answer is:
$\frac{A}{\sqrt{2}}$
Potential energy $=\frac{1}{2} k x^2$
Kinetic energy $=\frac{1}{2} k A^2-\frac{1}{2} k x^2$
According to given condition
$\frac{1}{2} k x^2=\frac{1}{2} k A^2-\frac{1}{2} k x^2$
$k 2 x^2=k A^2$
$x^2=\frac{A^2}{2}$
$x=\frac{A}{\sqrt{2}}$
Kinetic energy $=\frac{1}{2} k A^2-\frac{1}{2} k x^2$
According to given condition
$\frac{1}{2} k x^2=\frac{1}{2} k A^2-\frac{1}{2} k x^2$
$k 2 x^2=k A^2$
$x^2=\frac{A^2}{2}$
$x=\frac{A}{\sqrt{2}}$
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