A little platform oscile with a frequency of 4hz and an amplitude of 7 cm, attached to a vertical spring. A small object is poised on the top of the spring in the exactly moment it is in the lowest position. Admit the object is so light that it doesn't change the oscillation perceptively.
A.what is the distance from the equilibrium position of the platform that the object loses contact?
B.the velocity in this instant
(sorry for my bad English. If anyone want to reformulate the question for better understanding, it would be welcome)
Stacey Warren - Expert brainly.com
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*a small object is poused/put
at the highest point when the platform is returning to equilibrium.
at highest point velocity of both object and platform will be zero but acceleration of platform will be greater than the object hence it will lose contact there.
Actually, the object will ever be under gravitational force, so I think it will lose contact while the platform is already going down. (the answer isn't something like Xo+7)
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going down????? but you put it when the platform was at its lowest position
Remember, once the platform and object pass the equilibrium position on the way up, they are slowing down. Once the acceleration of the platform is <9.8 m/s^2, the object will lose contact with the platform.
I understood it. Intuitively, it seems that the object will lose contact at the top. But the spring realizes a force in down direction at this moment, so the acceleration of the object gets higher to this down direction. That's why the platform has zero velocity in amplitude points. Zero velocity at the top, and then there comes a time the platform gains so much velocity that it is way above the object falling velocity.
I guess that's it (I already have the answer, but can't get to that)
1.58 cm above the equilibrium point. (answer)
The amplitude is 7 cm.