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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 energiesat 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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a

32R

b

R2

c

3 R

d

2 R

answer is C.

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Detailed Solution

When the particle crosses D, its speed is half the maximum speed∴ V=vmaxRR2−x2 or  vmax2=vmaxRR2−x2 or x=32R∴  Distance BD=2x=3R
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