Question

Consider a two-mass-spring flexible mechanical system given below. Mass 1 Mass 2 In the system, u(t) is the input force, k = 1 is the spring constant, #1 and x2 are, respectively, the displacements of Mass 1 and Mass 2, which have masses of mı = m2 = 1. Assume that there is no friction on the surfaces. (a) Drive a differential equation of the mechanical system in terms of the displacement of Mass 2, i.e. X2

How can we do the same thing but for the mass 1?

Thank you!

Answer 1

x K m2 mi We have to derive a differential equation of the mechanical System in terms of the displacement of Mass 1.i.ex, Total kinetic energy of the system (T) T = 1 m, 2,2 + 2m₂ x ² Total potential enesgy of the system (v) v= K (12-21)² Using Lagrange's eqn of motion ƏT dv (07) - + where i = 1,2- Substituting the value of T, V and Fi we get + m2 22² e (4 mi, 2 + {m, xiz) dt lor dre ---n d dt = Fi dxi dxi de [mpe Jx 08, (maxi) - 0 + k (42.21) = 4 or, maxi + Khg_kxq = u - For i=1 we have m, x; tkx, - kaz=0 or mx tkx, = KX2 or, x2 = 1 / (m, ž + k*,) - ③

Now differentiating equ ® both sides with respect to t. d3x, dt3 tk da, dt again differentiating 2 = 1/2 (m, dox + k d²x, + k dty dtz Now substitute these values in eqn o we get dx ho tot + km, do Mai mi tkd ²x dty de ou + Kx;) - kx, = u or, m, m2 dx, K + m2 =u d²x1 dtz tm. d²x1 dt2 dty 08 m, m2 da, dty mimi +(m, tma) d²2 dt? Now for m = m₂ = 1, k=1 We have required differential eqn in terms of the displacement mass 1 is dya, + 2d²x1 =U. dty dt²