What is the angular frequency of oscillations in a SHM system with weight?

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SUMMARY

The discussion focuses on calculating the angular frequency of oscillations in a simple harmonic motion (SHM) system involving a mass hanging on a spring. The equilibrium displacement due to gravity is given as 0.1m, leading to the conclusion that the angular frequency, denoted as ω, is 9.9 s-1. The participants emphasize the importance of equating forces at equilibrium and utilizing the equation of motion, specifically d²x/dt² + ω²x = g, to derive the solution.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with differential equations
  • Knowledge of force equilibrium in mechanical systems
  • Basic concepts of mass-spring systems
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  • Study the derivation of angular frequency in mass-spring systems
  • Learn about the integrating factor method for solving differential equations
  • Explore the concept of equilibrium in mechanical systems
  • Investigate the relationship between force, mass, and acceleration in SHM
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to simple harmonic motion and mass-spring systems.

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Homework Statement



A mass hanging on a spring is oscillating with SHM. The initial displacement due to gravity is 0.1m By considering the forces acting on the mass at equilibrium, calculate the angular frequency of the oscillations.

Homework Equations



[tex]\frac{d^2x}{dt^2} + {w}^{2}x = g[/tex]

The Attempt at a Solution



I used the integrating factor method to get

[tex]x = \frac{m}{kg} (1-{e}^{-kt/m} -0.1{e}^{-kt/m}[/tex]

I'm not really sure where to go from there

I always get stuck on questions when they have weight!
 
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okay dw i got it now, 9.9 s-1 you just equate the forces at equilibrium with x = 0.1
 
$$\frac{d^2x}{dt^2} + {w}^{2}x = g$$

The Attempt at a Solution



I used the integrating factor method to get

$$ x = \frac{m}{kg} (1-{e}^{-kt/m} -0.1{e}^{-kt/m}) $$

I'm not really sure where to go from there
... I think you are getting ahead of yourself.
The weight adds an extra force into your free-body diagram.

If you have positive displacements downwards, measured from the unstretched length position,
then ##-ky+mg=m\ddot y## (you don't need to solve it!)
... divide through by m gets the same relation you have - see what ##\omega^2## is equal to?

You don't know k or m.
But you do know what the equilibrium displacement is
- what is special about the forces at equilibrium?
 

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