Question

A coin is placed on a horizontal platform which undergoes vertical shm of angular displacement w. the amplitude of oscillation is gradually increased. the coin will leave contact with the platform for the 1st

Answer 1

Magnitude of Acceleration of particle in SHM is given by
|a| = ω²y , where y is the displaced of particle
for maximum acceleration , y will be max
displacement of particle from mean position be max , then it is known as Amplitude . so, maximum acceleration , $\bold{a_{max}=\omega^2A}$

Now, when coin leave the contact with the platform for the 1st time ,
Then, weight of coin = force due to SHM motion at maximum displacement { y = A}
mg = mω²A⇒ A = g/ω²

Hence, amplitude of coin is g/ω²