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(C) The total energy of a body in simple harmonic motion is $E = \frac{1}{2} k A^2,$ where $A$ is the amplitude.Given that $AE = 2R$ and $C$ is the mi

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$\sqrt{3} R$

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(C) The total energy of a body in simple harmonic motion is $E = \frac{1}{2} k A^2,$ where $A$ is the amplitude.Given that $AE = 2R$ and $C$ is the midpoint,the amplitude $A = R$.The kinetic energy at any position $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$.According to the problem,at $B$ and $D$,the kinetic energy is one-fourth of the maximum value:$\frac{1}{2} k (A^2 - x^2) = \frac{1}{4} (\frac{1}{2} k A^2)$$A^2 - x^2 = \frac{1}{4} A^2$$x^2 = \frac{3}{4} A^2$$x = \pm \frac{\sqrt{3}}{2} A$Since $A = R$,the positions of $B$ and $D$ are $x = \pm \frac{\sqrt{3}}{2} R$.The distance between $B$ and $D$ is $|x_D - x_B| = |\frac{\sqrt{3}}{2} R - (-\frac{\sqrt{3}}{2} R)| = \sqrt{3} R$.

$A$ body performs simple harmonic oscillations along the straight line $ABCDE$ with $C$ as the midpoint of $AE.$ Its kinetic energies at $B$ and $D$ are each one-fourth of its maximum value. If $AE = 2R,$ the distance between $B$ and $D$ is

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