anonymous
  • anonymous
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)
Physics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
*a small object is poused/put
anonymous
  • anonymous
at the highest point when the platform is returning to equilibrium. |dw:1345657752597:dw| 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.
anonymous
  • anonymous
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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
going down????? but you put it when the platform was at its lowest position
NoelGreco
  • NoelGreco
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.
anonymous
  • anonymous
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)
anonymous
  • anonymous
1.58 cm above the equilibrium point. (answer) The amplitude is 7 cm.

Looking for something else?

Not the answer you are looking for? Search for more explanations.