A particle is executing simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/sec. The distance of the particle from the mean position when its speed becomes 5 cm/s is

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3 cm

22 cm

23 cm

32 cm

answer is C.

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Detailed Solution

v = ωa2-y2At y = 0, vmax = 10 cm/sAs a = 4 cm, vmax = ωa hence ω = 2.5 rad/sHence, 5 = 2.516-y2or y = 23 cm

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