Physics - SHM Question with Solution | TestHub

In vertical simple harmonic motion, maximum acceleration$\left(\mathit{\text{a}}{\omega }^2\right)$and so the maximum force$\left(\mathit{\text{m}}\mathit{\text{a}}{\omega }^2\right)$will be at extreme positions. At highest position, force will be towards mean position and so it will be downwards. At lowest position, force will be towards mean position and so it will be upwards. This is opposite to weight direction of the coin. The coin will leave contact with the platform for the first time when$\mathit{\text{m}}\left(\mathit{\text{a}}{\omega }^2\right)\ge \mathit{\text{m}}\text{g}$at the lowest position of the platform.