Question

A coin is placed on a horizontal platform which undergoes vertically simple harmonic motion of Angular frequency ω . the amplitude of oscillation is gradually increased . the coin will leave contact with the platform for the first time .[AIEEE 2006]
(a) at mean position of the platform
(b) for an amplitude of g/ω²
(c) for an amplitude of g/ω²
(d) at the heighest position of the platform ​

Answer 1

Answer:

The coin will leave contact at any point when the acceleration of platform is more than 'g' and in the same direction of 'g'. At the highest point, both of the conditions are satisfied. Acceleration at the highest point is equal to ω^2A, so we have

ω^2A≥g

=>A≥g/ω^2

HOPING THIS WILL HELP YOU PLZ MARK AS BRAINLIEST AND FOLLOW ME

Answer 2

Explanation :- As the amplitude is increased , the maximum acceleration of the platform (along with coin as long as they does not get separated increases)

If we draw the free body diagram of coin at one of the extreme positions as shown then from newton's law

mg - N = mω²A

For loosing contact with the platform

N = 0

so, A = g/ω² Answer

since option (B) /(c) is correct