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A coin is placed on a horizontal platfor...

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

C

for an amplitude of `(g)/(omega^2)`

D

for and amplitude of `sqrt((g)/(omega))`

Text Solution

Verified by Experts

The correct Answer is:
A, C


Let O be the mean position and a be the acceleration at a displacement x from O
At position I, `N-mg=ma`
`Nne0`
At position II, `mg-N=ma`
For `N=0` (lost of contant), `g=a=omega^2x`.
Loss of contact will occur for amplitude `x_(max)=(g)/(omega^2)` at the highest point of the motion.
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Knowledge Check

  • In the previous question, the angular frequency of the simple harmonic motion is omega . The coefficient of friction between the coin and the platform is mu . The amplitude of oscillation is gradually increased. The coin will begin to slip on the platform for the first time (i) at the extreme positions of oscillations (ii) at the mean position (iii) for an amplitude of (mug)/(omega^2) (iv) for an amplitude of (g)/(muomega^2)

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