MT-1620 al.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
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
MT-1620 al.2002 Return to the simplest system the single spring-mass This is a one degree-of-freedom system with the governing equation mg t q F First consider Free Vibration Set f=o resulting in mq+kq=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 g() mppt kepl=o =mp+k=0(in order to hold for all t Paul A Lagace @2001 Unit 20-2
qt 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: m q˙˙ + k q = F First consider… Free Vibration ⇒ Set F = 0 resulting in: m q˙˙ + 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 pt ⇒ m p2e pt + k e pt = 0 ⇒ 2 mp + k = 0 (in order to hold for all t) Paul A. Lagace © 2001 Unit 20 - 2
MT-1620 al.2002 k →P m Where a natural frequency of single m degree-of-freedom system [rad/sec stiffness Important concept that natural frequency mass So have the equation q()=C +I0 t +ce Paul A Lagace @2001 Unit 20-3
qt it 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: () = C1 e+ ω + C2 e− ω Paul A. Lagace © 2001 Unit 20 - 3
MT-1620 Fall 2002 from mathematics know this is (0)=C sinat C cost general solution Now use the initial conditions 0q=9→C q g0 This results in g(t)= q sinat go cost ith This is the basic, unforced response of the system So if one gives the system an initial displacement a and then lets go Paul A Lagace @2001 Unit 20-4
qt MIT - 16.20 Fall, 2002 from mathematics, know this is: q t() = C1 ′sinω t + C2 ′ cosω t general solution Now use the Initial Conditions: @ t = 0 q = q0 ⇒ C2 ′ = q0 @ t = 0 q˙ = q˙0 ⇒ C1 ′ = q ˙ 0 ω This results in: () = q ˙ 0 sinω t + q0 cosω 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: q0 = A q ˙ 0 = 0 Paul A. Lagace © 2001 Unit 20 - 4
MT-1620 al.2002 The response is q(t)=Acosot Figure 20.1 Basic unforced dynamic response of single spring-mass system A e 七 But, generally systems have a force, so need to consider Paul A Lagace @2001 Unit 20-5
qt MIT - 16.20 Fall, 2002 The response is: () = A cosω 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