# Solving Oscillations: Bead on a Block & Spring

• turdferguson
In summary, the bead will lose contact with the block when the block has a downward acceleration greater than g, or when the sum of the forces on the bead is zero. The mass of the bead is negligible and does not affect the calculations.
turdferguson

## 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

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?

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.

## 1. How do I solve for the period of oscillation for a bead on a block and spring system?

The period of oscillation for a bead on a block and spring system can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass of the bead, and k is the spring constant. This formula assumes that there is no friction or damping in the system.

## 2. Can the amplitude of oscillation be changed in a bead on a block and spring system?

Yes, the amplitude of oscillation can be changed by adjusting the initial displacement of the bead from its equilibrium position. The greater the displacement, the larger the amplitude of oscillation will be.

## 3. How does the mass of the bead affect the period of oscillation in a bead on a block and spring system?

The period of oscillation is inversely proportional to the square root of the mass of the bead. This means that as the mass of the bead increases, the period of oscillation will decrease.

## 4. What is the relationship between the spring constant and the period of oscillation in a bead on a block and spring system?

The period of oscillation is directly proportional to the square root of the spring constant. This means that as the spring constant increases, the period of oscillation will also increase.

## 5. How does friction or damping affect the motion of a bead on a block and spring system?

Friction and damping will cause the amplitude of oscillation to decrease over time, eventually leading to the bead coming to rest. This is because these factors work against the restoring force of the spring, reducing the amplitude of the oscillations. The presence of friction and damping can also change the period of oscillation, making it longer than the ideal period calculated using the formula mentioned in the first question.

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