NTA Abhyas JEE Main2020PhysicsOscillationsPractice
A horizontal platform, with an object placed on it, is executing SHM in the vertical direction. The amplitude of oscillation is 2.5 cm . Let the least period of these oscillations, so that the object is not detached, be equal to π 2 a . Then, the value of a is (Given: g = 10 m s - 2 )
Correct answer
5
Step-by-step solution
The object would not leave the platform when a m a x < g ∴ w 2 A < g ∴ w < 10 2.5 × 10 - 2 = 10 4 25 = 100 5 = 20 ∴ 2 π T < 20 ⇒ T > π 10