A vibration platform oscillates up and down with an amplitude of 11.8 cm at a controlled variable frequency.?

A vibration platform oscillates up and down with an amplitude of 11.8 cm at a controlled variable frequency. Suppose a small rock of unknown mass is placed on the platform. At what frequency will the rock just begin to leave the surface so that it starts to clatter?A vibration platform oscillates up and down with an amplitude of 11.8 cm at a controlled variable frequency.?
The platform would have to go down faster than it would under the acceleration of gravity.





For gravity, the time to go from the max to the trough would be;





distance = 1/2 acel * time^2





11.8 X 10^-2 m = 1/2 (9.81 m/s/s)*t^2





solve for t and you get t= 0.115 sec





That would be half a cycle. A complete cycle would be t= 0.31 seconds.





The frequency would be 3.2 hz. Anything faster would have the rock leave the platform.

beauty