question_answer

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     AIEEE  Solved  Paper-2006

A) at the mean position of the platform

B) for an amplitude of \[g/{{\omega }^{2}}\]

C) for an amplitude of\[{{g}^{2}}/{{\omega }^{2}}\]

D) at the highest position of the platform

Correct Answer: B

Solution :

As the amplitude is increased, the maximum acceleration of the platform (along with coin as long as they doesn't get separated) increases. If we draw the FBD for coin at one of the extreme positions as shown                 \[\left[ \begin{align}   & \therefore a={{\omega }^{2}}A \\  & in\,SHM \\ \end{align} \right]\] Then,   from   Newton's   second   law, \[mg-N=m{{\omega }^{2}}A\]. For loosing contact with the platform,\[N=0\] So,        \[A=g/{{\omega }^{2}}\]