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for a given omega, show that the position (or the envelope of the position) goes to zero quickest for a critically damped harmonic oscillator.

2007-12-11 16:10:32 · 2 answers · asked by Johnny 1 in Science & Mathematics Physics

2 answers

Well, that's the definition of critical damping. That is - when an oscillator is damped in such a way that it reaches zero and stays there in the least possible time, that's called "critical damping."

2007-12-11 16:32:50 · answer #1 · answered by jgoulden 7 · 0 0

A mass of a million.sixty 4 kg stretches a vertical spring 0.315 m. If the spring is stretched one greater desirable 0.138 m and released, how long does it take to attain the (new) equilibrium place decrease back? 1st we would prefer to ensure the pliability consistent for this spring. F = ok * d F = weight of merchandise = a million.sixty 4 * 9.8 = sixteen.072 N d = distance stretched = 0.315 m sixteen.072 = ok * 0.315 ok = 51.0222 N/m The time required for the item to make a million finished oscillation is the era. The era is self sufficient of the section the spring is stretched. era = 2 * ? * (mass ÷ ok)^0.5 era = 2 * ? * (a million.sixty 4 ÷ 51.0222)^0.5 = a million.126 seconds

2016-10-11 02:51:26 · answer #2 · answered by ? 4 · 0 0

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