Question:

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

Updated On: Jul 5, 2022
  • at the mean position of the platform
  • for an amplitude of $ g/\omega^{2} $
  • for an amplitude of $ \sqrt{g/\omega} $
  • at the highest position of the platform
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The Correct Option is B

Solution and Explanation

As the amplitude is increased, the maximum acceleration of the platform (along with coin as long as they doesn't get separated) increases.
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If we draw the FBD for coin at one of the extreme positions as shown then from Newton's law, $m g-N=m \omega^{2} A$ For loosing contact with the platform, $N=0$ So, $\quad A=g / \omega^{2}$
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