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A body performs SHM along the straight line segment ABCDE with C as the mid point of segment AE (A and E are the extreme position for the SHM). Its kinetic energies
at B and D arc each one fourth of its maximum value. If length of segment AE is 2R, then the distance between B and D is

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$\frac{\sqrt{3}}{2} R$

$\frac{R}{\sqrt{2}}$

$\sqrt{3} R$

$\sqrt{2} R$

answer is C.

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Detailed Solution

When the particle crosses D, its speed is half the maximum speed

$∴ V = \frac{v_{max}}{R} \sqrt{R^{2} - x^{2}}$

$or \frac{v_{max}}{2} = \frac{v_{max}}{R} \sqrt{R^{2} - x^{2}} or x = \frac{\sqrt{3}}{2} R$

$∴ Distance BD = 2 x = \sqrt{3} R$

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