ch3sol-to-class (1)

ch3sol-to-class (1) - 6 CHAPTER 3 SDOF VIBRATION: WITH...

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6 CHAPTER 3 SDOF VIBRATION: WITH DAMPING 4. For a mass-spring-damper system in free vibration: (a) Derive the equation of motion and solve for the transient response x ( t ) that is driven by the initial conditions x (0) = x 0 and _ x (0) = v 0 . (b) Assume parameter values: k = 1 lb/in, weight W = 100 lb, x 0 = 0 , v 0 = 10 in/s. Vary the damping constant c so that both underdamped and overdamped responses can be demonstrated. Try values of c such that the cases = 0 : 1 and = 0 : 9 are obtained. (c) For the case above with = 0 : 1 , vary initial velocity v 0 (zero displacement) as a function of initial velocity. Solution: (a) The equation of motion for a free mass-spring-damper system is given by m x + c _ x + kx = 0 or x + 2 n _ x + ! 2 n x = 0 : We assume a solution of the form x ( t ) = A exp( rt ) or x ( t ) = A cos rt + B sin rt: _ x and x; A exp( rt )[ mr 2 + cr + k ] = 0 : Since neither A nor the exponential equal zero, the polynomial in the brackets must equal zero, mr 2 + cr + k = 0 : This second order polynomial is the characteristic equation for the system and it can be solved for the two values of r using the quadratic equation, r 1 ; 2 = c ± p c 2 4 mk 2 m : The form of the solution changes depending on the value of c 2 4 mk: Overdamped Response: For c 2 4 mk > 0 (equivalently 1) ; the system is overdamped, and the solution takes the form x ( t ) = Ae + p 2 1 ± ! n t + Be p 2 1 ± ! n t : Upon substituting the initial conditions, the constant coe¢ cients of integration are found to be A = 1 2 x o + v o + o ! n ! n p 2 1 ! and B = 1 2 x o v o + o ! n ! n p 2 1 ! : Critically Damped Response: For c 2 4 mk = 0 (equivalently = 1) ; the system is critically damped, and the solution takes the form x ( t ) = ( A + Bt ) exp ( ! n t ) ;
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7 where the constant coe¢ cients of integration are found to be A = x 0 and B = v 0 + x 0 ! n : Underdamped Response: For c 2 4 mk < 0 (equivalently 0 1) , the system is underdamped, and the solution takes the form x ( t ) = exp ( n t ) ( A cos ! d t + B sin ! d t ) ; where ! d = ! n p 1 2 . Upon substituting the initial conditions, the constant coe¢ cients of inte- gration are found to be A = x 0 and B = v 0 + x 0 n ! d : An alternate form of the solution for the underdamped system is x ( t ) = exp ( n t ) [ C cos ( ! d t ± )] : x 0 = C cos ± and v 0 = C ( n cos ± + ! d sin ± ) : The amplitude and the phase, C and ±; are found to be C = s x 2 0 + v 0 + &!
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ch3sol-to-class (1) - 6 CHAPTER 3 SDOF VIBRATION: WITH...

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