Q. 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
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
Explanation
(b) As the amplitude is increased, the maximum acceleration of the platform (along with coin as long as they doesn’t get separated) increases
[image:5_image258.emf]
if we draw the FBD for coin at the extreme position as shown
[image:5_image259.emf]
then from Newton’s law, $mg - N = m\ \omega^{2}A$
For loosing contact with the platform, $N = 0$
$So,\ \ A = \frac{g}{\omega^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.