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Q.

A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

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a

at the mean position of the platform

b

for an amplitude of g/ω2

c

for an amplitude of g2/ω2

d

at the highest position of the platform

answer is B.

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Detailed Solution

The free body diagram for the coin at one of the extreme position is shown in fig.According to Newtons law, mg−N=mω2AFor loosing contact with platform, N = 0∴ mg=mω2A  or  A=g/ω2
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