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unit20 - MIT 16.20 Fall 2002 Unit 20 Solutions for Single...

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MIT - 16.20 Fall, 2002 Unit 20 Solutions for Single Spring-Mass Systems Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001

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q t MIT - 16.20 Fall, 2002 Return to the simplest system: the single spring-mass… This is a one degree-of-freedom system with the governing equation: mq ˙˙ + k q = F First consider… Free Vibration Set F = 0 resulting in: mq ˙˙ + k q = 0 The solution to this is the homogeneous solution to the general equation. For an Ordinary Differential Equation of this form, know that the solution is of the form: ( ) = e p t m p 2 e p t + ke p t = 0 2 m p + k = 0 (in order to hold for all t) Paul A. Lagace © 2001 Unit 20 - 2
q t i t i t MIT - 16.20 Fall, 2002 2 k p = m p = ± i k m where: i = 1 k m = ω = natural frequency of single degree-of-freedom system [rad/sec] Important concept that natural frequency = stiffness mass So have the equation: ( ) = C 1 e + ω + C 2 e ω Paul A. Lagace © 2001 Unit 20 - 3

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q t MIT - 16.20 Fall, 2002 from mathematics, know this is: q t ( ) = C 1 sin ω t + C 2 cos ω t general solution Now use the Initial Conditions: @ t = 0 q = q 0 C 2 = q 0 @ t = 0 q ˙ = q ˙ 0 C 1 = q ˙ 0 ω This results in: ( ) = q ˙ 0 sin ω t + q 0 cos ω t ω with: ω = k m This is the basic, unforced response of the system So if one gives the system an initial displacement A and then lets go: q 0 = A q ˙ 0 = 0 Paul A. Lagace © 2001 Unit 20 - 4
q t MIT - 16.20 Fall, 2002 The response is: ( ) = A cos ω t Figure 20.1 Basic unforced dynamic response of single spring-mass system But, generally systems have a force, so need to consider: Paul A. Lagace © 2001 Unit 20 - 5

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MIT - 16.20 Fall, 2002 Forced Vibration The homogeneous solution is still valid, but must add a particular solution The simplest case here is a constant load with time… Figure 20.2 Representation of constant applied load with time (think of the load applied suddenly step function at t = 0) The governing equation is: mq ˙˙ + k q = F 0 The particular solution has no time dependence since the force has no time dependence: F 0 q particular = k Paul A. Lagace © 2001 Unit 20 - 6
q t MIT - 16.20 Fall, 2002 Now use the homogeneous solution with this to get the total solution: F ( ) = C 1 sin ω t + C 2 cos ω t + 0 k The Initial Conditions are: @ t = 0 q = 0 q ˙ = 0 0 q ( ) = 0 C = F 0 2 k q 0 ˙ ( ) = 0 C 1 = 0 So the final solution is: F q = 0 ( 1 cos ω t ) k with ω = k m Plotting this: Paul A. Lagace © 2001 Unit 20 - 7

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MIT - 16.20 Fall, 2002 Figure 20.3 Ideal dynamic response of single spring-mass system to constant force Dynamic response
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unit20 - MIT 16.20 Fall 2002 Unit 20 Solutions for Single...

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