Q136 / 150 · JEE Main 2006 · Physics

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
  • Correct answer
    C
    for an amplitude of $\frac{\mathrm{g}}{\omega^2}$
  • D
    for an amplitude of $\frac{\mathrm{g}^2}{\omega^2}$

Solution

$A \omega^2=g$ $\Rightarrow A=g / \omega^2$

Hence, the correct answer is option: 'C'.

Relevant Questions