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