Overshoot of 15% what is damping ratio

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SUMMARY

The discussion focuses on calculating the damping ratio (##\zeta_0##) required for a shock absorber to limit its overshoot to 15% of its initial displacement. The log decrement formula, ##\delta = \ln\frac{x_1}{x_2} = \frac{2\pi\zeta}{\sqrt{1 - \zeta^2}}##, is central to the calculations. Participants confirm that for a 15% overshoot, the ratio ##\frac{x_1}{x_2}## should be set to 1.15 to solve for ##\zeta##. Clarifications on the definition of overshoot and the correct amplitudes for comparison during oscillation are also discussed.

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  • Understanding of damping ratios in mechanical systems
  • Familiarity with the log decrement formula
  • Basic knowledge of oscillatory motion and overshoot concepts
  • Ability to solve algebraic equations involving logarithms
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  • Study the derivation and applications of the log decrement formula
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  • Learn about different types of damping (underdamped, critically damped, overdamped)
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Mechanical engineers, control system designers, and students studying dynamics and vibration analysis will benefit from this discussion.

Dustinsfl
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Homework Statement


A shock absorber is to be designed to limit its overshoot to ##15## percent of its initial displacement when released. Find the damping ratio ##\zeta_0## required.

Homework Equations

The Attempt at a Solution



My question is since log decrement, ##\delta = \ln\frac{x_1}{x_2} = \frac{2\pi\zeta}{\sqrt{1 - \zeta^2}}##, for a 15% overshoot, do I set ##\frac{x_1}{x_2} = 1.15## and then solve for ##\zeta##?
 
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How is overshoot defined? In the way you use it it looks odd (1000% overshoot would mean a strong damping then).
Also make sure you use the correct amplitudes to compare - after 1/2 of the oscillation or a full oscillation?
 
Dustinsfl said:
My question is since log decrement, ##\delta = \ln\frac{x_1}{x_2} = \frac{2\pi\zeta}{\sqrt{1 - \zeta^2}}##, for a 15% overshoot, do I set ##\frac{x_1}{x_2} = 1.15## and then solve for ##\zeta##?

yes
 

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