Frequency of Oscillation for a Spring with Attached Block

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Homework Help Overview

The problem involves a spring system with a block attached, focusing on the frequency of oscillation after the block is displaced from its equilibrium position. The context includes concepts from mechanics and oscillatory motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the mass at equilibrium and the relationship between the spring constant and mass. Questions arise about the correct application of Hooke's law and the interpretation of energy versus force.

Discussion Status

The discussion has progressed with participants clarifying the relationship between the spring force and gravitational force. Some guidance has been offered regarding the correct expressions to use, and there is a growing understanding of the concepts involved.

Contextual Notes

There is a mention of missing mass information in the problem statement, which participants are addressing through their discussions on force relationships and spring constants.

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Homework Statement



A spring is hung from the ceiling. When a block is attached to its end, it streches 19.3 cm before reaching its new equilibrim length. The block is then pulled down slightly and released. What is the frequency of the oscillation?

Homework Equations


f =[tex]\frac{\omega}{2\pi}[/tex]

[tex]\omega[/tex]= [tex]\sqrt{\frac{k}{m}}[/tex]

The Attempt at a Solution


i have no idea how to start this question and what to make of the mention of the equilibrium length pls. help
 
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Which two opposite and equal forces act upon the mass when the spring is in its equilibrium position?
 
the weight and the normal force but they don't mention any mass in the question
 
You don't want to calculate the mass you want to calculate k/m. The restoring force exerted by the spring is equal to the weight. Can you write this down in formulaic form?
 
wld it be .5*k*x^2= mg where K/m = 2g/x^2 ??
 
That is not Hooke's law. You used the elastic potential energy as a force which is not true because it is an energy. Find the correct expression for the restoring force exerted by the spring.
 
oh do u mean F=kx so that wld be kx=mg where k/m = g/x
 
Yep.
 
thanx i get it now :)
 

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