Page 1 of 1

7. As we know from Section 20, the forced vibrations of an undamped spring-mass system are described by the differential

Posted: Wed Jul 06, 2022 11:50 am
by answerhappygod
7 As We Know From Section 20 The Forced Vibrations Of An Undamped Spring Mass System Are Described By The Differential 1
7 As We Know From Section 20 The Forced Vibrations Of An Undamped Spring Mass System Are Described By The Differential 1 (24.37 KiB) Viewed 13 times
7 As We Know From Section 20 The Forced Vibrations Of An Undamped Spring Mass System Are Described By The Differential 2
7 As We Know From Section 20 The Forced Vibrations Of An Undamped Spring Mass System Are Described By The Differential 2 (51.71 KiB) Viewed 13 times
7 As We Know From Section 20 The Forced Vibrations Of An Undamped Spring Mass System Are Described By The Differential 3
7 As We Know From Section 20 The Forced Vibrations Of An Undamped Spring Mass System Are Described By The Differential 3 (29.4 KiB) Viewed 13 times
7. As we know from Section 20, the forced vibrations of an undamped spring-mass system are described by the differential equation Mx"+kx = f(t), where x() is the displacement and f(t) is the impressed external force or "forcing function." If x(0) = x'(0) = 0, find the functions A(f) and h(t) and write down the solution x(t) for any f(t).
y" +ay' + by = f(t) that satisfy the initial conditions y (0) = y'(0) = 0 describing a mechanical or electrical system at rest in its equilibrium posi- tion. The input f(t) can be thought of as an impressed external force F or elec- tromotive force E that begins to act at time t = 0, as discussed in Section 20. When this input is the unit step function u(t) defined in Problem 49-2(a), the solution (or output) y(t) is denoted by A(t) and called the indicial response; that is, SO A" + aA' + bA= u(t). By applying the Laplace transformation L and using formulas (3) and (4) in Section 50, we obtain p²L[A] + apL[A] + bL[A] = [[u(t)] = 1/ L[A]= (5) 1 1 p p²+ap+b 11 pz(p)'
[[h(t)]= 1 z(p) (14)
k 7. A(I) = (1-006 √7) 1-cos -t k M 1 x(t) = √√k. So f Mk Jo fff(t) sin k M 1 h(t) = sin -t, √Mk k VM -(t-t)dt.