A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω.

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asked Mar 23, 2018 in Physics by paayal (26,720 points) 4 6 11

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
(a) at the highest position of the platform
(b) at the mean position of the platform
(c) for an amplitude of g/ω2

(d)  for an amplitude of g2/ω2

1 Answer

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answered Mar 23, 2018 by sanjaydas (61,430 points) 5 7 7
selected Mar 24, 2018 by Vikash Kumar
 
Best answer

(c) : In vertical simple harmonic motion, maximum acceleration (aω2) and so the maximum force (maω2) will be at extreme positions. At highest position, force will be towards mean position and s o it will be downwards. At lowest position, force will be towards mean position and so it will be upwards. This is opposite to weight direction of the coin. The coin will leave contact will the platform for the first time when m(aω2≥ mg at the lowest position of the platform.

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