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

(C) for an amplitude of $\frac{\mathrm{g}}{\omega^{2}}$

(A) at the highest position of the platform

(B) at the mean position of the platform

(D) for an amplitude of $\frac{g^{2}}{\omega^{2}}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Elasticity & SHM

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

Answer: (C)

The platform can only push up on the coin, so the largest downward acceleration it can impart while they stay in contact is

g

, reached when the normal force just falls to zero. In vertical simple harmonic motion the peak acceleration is

ω^{2} A

, directed downward at the top of the swing. Contact is first lost when

ω^{2} A = g

, that is for an amplitude

A = \frac{g}{ω ^{2}}

, and it happens at the highest point of the motion.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option