Chapter3Section4

# Chapter3Section4 - 3.4 Mechanical Vibrations Mass Spring Dashpot c m x Equilibrium Point Set-up Attach a mass m to a spring and a dashpot x t is

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3.4: Mechanical Vibrations ± Mass - Spring - Dashpot x Equilibrium Point m c Set-up: Attach a mass, m , to a spring and a dashpot. x H t L is the distance from equilibrium. If x H t L > 0, then the spring is stretched. If x H t L < 0, then the spring is compressed. A d a s h p o t is a piston that always opposes motion. Assume that the force of opposition is proportional to velocity with constant of proportionality c > 0. Hence, we have F R = - c v = - c dx dt . If c = 0, then the motion is u n d a m p e d . If c 0, then the motion is d a m p e d . Hooks Law : The restorative force, F S , is proportional to the distance stretched and opposes motion. So, for k > 0, F S = - k x . External Force F E = F H t L . If F E = F H t L = 0, then the motion is f r e e . If F E = F H t L 0, then the motion if f o r c e d . Newton's Law : F T = m a = m d 2 x dt 2 . Total Forces on Mass : F T = F S + F R + F E . Hence, we need to solve m d 2 x dt 2 + c dx dt + k x = F E = F H t L .

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± Free Undamped Motion: F H t L = 0, c = 0. Then, we need to solve the basic second order, linear, homogeneous differential equation with constant coefficients: m d 2 x dt 2 + k x = 0 ± d 2 x dt 2 + k m x = 0 ± d 2 x dt 2 + Ω 0 2 x = 0, where we define Ω 0 = k m . Thus, we have a characteristic equation of r 2 + Ω 0 2 = 0 ± r = –Ω 0 i . Thus, our solution can be written as a linear combination of sines and cosines, x H t L = A cos H Ω 0 t L + B sin H Ω 0 t L = C A A C cos H Ω 0 t L + B C sin H Ω 0 t LE , where C = A 2 + B 2 = C @ cos H Α L cos H Ω 0 t L + sin H Α L
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## This note was uploaded on 01/13/2012 for the course MATH 306 taught by Professor Keithemmert during the Spring '11 term at Tarleton.

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Chapter3Section4 - 3.4 Mechanical Vibrations Mass Spring Dashpot c m x Equilibrium Point Set-up Attach a mass m to a spring and a dashpot x t is

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