Linear Simple Harmonic Oscillator: period a direct linear proportion to mass?

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SUMMARY

The discussion clarifies that in a simple harmonic oscillator, the period is not directly proportional to mass but rather proportional to the square root of mass. The relationship is defined by the formula T = 2π√(m/k), where T represents the period, m is the mass, and k is the spring constant. As mass increases, the period increases, but this increase is not linear; it follows a square root function. Therefore, doubling the mass does not double the period but increases it by a factor of √2.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with the spring constant (k) and its role in oscillation
  • Basic knowledge of angular frequency and its calculation
  • Ability to manipulate algebraic equations involving square roots
NEXT STEPS
  • Study the derivation of the period formula for simple harmonic oscillators
  • Explore the effects of varying the spring constant (k) on oscillation
  • Investigate real-world applications of simple harmonic motion in engineering
  • Learn about damping effects on simple harmonic oscillators and their impact on period
USEFUL FOR

Students of physics, mechanical engineers, and anyone interested in the dynamics of oscillatory systems will benefit from this discussion.

charlatain
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If a mass that hangs suspended vertically from a spring is increased, then won't the period increase as a direct linear proportion? (Because the larger mass has a greater inertia and will require a larger force and longer time to change the direction of motion on each oscillation?)

Some guidance would be greatly appreciated!
 
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A simple harmonic oscillator with spring constant k and mass m has angular frequency \omega = \sqrt{k/m}, so the period is T = 2\pi/\omega = 2\pi\sqrt{m/k}.

Thus, period is proportional to the square root of the mass.
 

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