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:

A)  for an amplitude of \[2g/3{{\omega }^{2}}\]

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

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

D)  for an amplitude of \[2{{g}^{2}}/3{{\omega }^{2}}\]

Correct Answer: B

Solution :

As the amplitude is increased, the maximum acceleration of the platform (along with coil as long as they doesn't get separated increase. If we draw the FBD for coil at one of the extreme positions as shown along side, then from Newton's law,             \[mg-N=m{{\omega }^{2}}A\] For loosing contact with the platform N = 0 \[\Rightarrow \,mg=m{{\omega }^{2}}A\]           So, \[A=\frac{g}{{{\omega }^{2}}}\]