Solving Oscillations: Bead on a Block & Spring

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Homework Help Overview

The problem involves a block attached to a spring oscillating vertically, with a bead placed on top of the block. The discussion centers around determining the distance from the block's equilibrium position at which the bead loses contact with the block during its motion.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on the bead and the block, questioning whether the bead loses contact when the normal force or the sum of forces becomes zero. There is discussion about the role of gravity and the acceleration of the block in relation to the bead's motion.

Discussion Status

Participants are actively engaging with the problem, offering insights into the forces at play and the conditions under which the bead may lose contact. There is a recognition that the mass of the bead may not be necessary for finding a numerical answer, and some guidance has been provided regarding the relationship between acceleration and the forces acting on the bead.

Contextual Notes

There is an assumption that the bead's mass is negligible compared to the block, and the discussion reflects on the implications of this assumption for the forces involved. Participants are also considering the lack of specific mass values for the block and bead in their reasoning.

turdferguson
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Homework Statement


A block attached to a spring underneath oscillates vertically with a frequency of 4Hz and an amplitude of 7.00cm. A tiny bead is placed on top of the block jut as it reaches its lowest point. Assume the bead's mass is so small that its effect on the motion of the block is negligible. At what distance from the block's equilibrium position does the bead lose contact with the block


Homework Equations


Hooke's Law?


The Attempt at a Solution


My first instict was that the bead would launch up until the maximum amplitude when the spring starts to move downward. Then I thought about Hooke's Law. Am I supposed to sum up the forces and say the bead is launched when there is no normal (restoring) force? This would mean the bead loses contact at equilibrium. Am I right about this?
 
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Ive found a better way of explaning my thoughts. Will spring accelerate the block and the bead upward until the sum of the forces is zero, or until the normal (restoring) force is zero? I am pretty sure its the former, but without the mass of the block or bead, can I get a numerical answer?
 
Last edited:
I'm think it is until the normal force is zero. Since after the bead is launched into the air, the only force acting on it would be the force of gravity.
 
turdferguson said:
Ive found a better way of explaning my thoughts. Will spring accelerate the block and the bead upward until the sum of the forces is zero, or until the normal (restoring) force is zero? I am pretty sure its the former, but without the mass of the block or bead, can I get a numerical answer?

That's the right idea. The forces on the bead are its weight and the reaction force from the block. You know the motion of the bead (the same as the block) so you can find its acceleration and therefore the resultant force on it. The mass of the bead will cancel out in the equations.

The force between the bead and block can only act in one direction. The bead isn't glued to the block, so the block can only "push" not "pull".

You could also work in terms of acceleration not force. The maximum downwards acceleration the bead can have is g (caused by gravity). When the block has a bigger acceleration than g downwards, the bead will lose contact with it.
 

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