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,

• A. at the highest position of the platform
• B. at the mean position of the platform
• C. for an amplitude of $\frac{g}{{\omega }^2}$
• D. for an amplitude of $\sqrt{\frac{g}{\omega }}$

Answer 1

Answer: (A), (C)

The correct options are
A at the highest position of the platform
C for an amplitude of $\frac{g}{{\omega }^2}$


Let $O$ be the mean position and a be the acceleration at a displacement $x$ from $O$.
At lowest position $I$
$N-mg=ma$
$\therefore N\ne 0$
At topmost position $II$,
$mg-N=ma$
For $N=0$ (loss of contact),
$g=a={\omega }^{2}x$.
Loss of contact will occur for amplitude $x_{max}=\frac{g}{{\omega }^2}$ at the highest point of the motion.

Why this question? The coin will leave contact at any point when the acceleration of platform is more than ${}^{\prime }{g}^{\prime }$ and in the same direction of ${}^{\prime }{g}^{\prime }$. At the highest point, both of the conditions are satisfied.