Question

A team of researchers are working to find a solution to the anticipated global lack of ventilators. You are asked to solve a vibration problem with the prototype. The system has a mass m = 5 kg, and the moving parts induce a harmonic force F(t) = 57 cos (66 t) N. The damping is negligible, and the system is assumed to start from rest with no velocity. 1. Write the total response of the system x(t) as a function of only F0/k, ω, and ωn . 2. Find the amplitude of the motion in (1) as a function of only F0/k, ω, and ωn. Consider that cos(2A)-cos(2B) = -2 sin(A)sin(B) 3. Determine the minimum stiffness of the system to keep the maximum vibration lower than 1.5 mm. In the next version, it is desirable that the main mass of the ventilator experiences no vibration at all at the operating frequency ω. 4. Select the mass and stiffness of a dynamic vibration absorber so that its mass moves no more than 10 mm.

Answer 1

If a force FCE) = Focos (wt) acts on mass m of an undamped system, the equation of motion is matka = F coswt - ☺ This equation has : 1. A homogenous solution x lt) where - nolt) = C, costw t + çsin (unt) - 2 k in which wn= of the system and constants. is the natural frequency 6 and C are arbitrary . 2. A particular solution X (L) where xp (t) = x coswt - ② in which & denotes amplitude of nolt)

Since xolts satisfies equation 0 mip + knp = Fo cosut = - mw²x coscot & kx cosut - Fo coswt X - F 's Folk K-mw? => X = F/K The total response alt) = n(t) + x, (t) R(t) = C, csont + C, sinunt + Fo loswt - 4 at tog x=0 x = 0 / Initial conditions) I given in question / -

. Using these initial conditions in (4) C, = . fo ; C₂ = 0 K-mw hence, the total response x(1) is CoswE (2) . Let us substitute (Folx I with X in alt) cabo we now have X(t) = -X cos wat + X wswt = x ( cos wt - cos wat ) --x ( 2 sin but sin cant) 2 sin futisin Casertin 28 voting for using I los 2A - cos 28-25inAsing)

- X(t)= -2& sinont) sint) x0) - 2 F/4 sin cant sin lut - 1 Amplitude of xlt) = 2 Fox - Since & = 57N w = 66 rad/s and m= 5kg and xlt) = -2 Folk sin wat sin fwt) ( from 6) we have amplitude of r(t) = X = 250/4 Xo can be re-written as 2 Fo K-mw?

Since we want X < 1.5mm, we want 1.5mm = 15 m L 2 F komor 1.5 1000 I =) k-mw² > 2000 to 1-5 ks 2000 Fo & mw- 1.5 =) k> 2000 x 57 + 5x 66² 1.5 - - k> 76000 + 21780 - Ik > 97.78 kN/m) This is the minimum stiffness of to keep amplitude lower than the system 1.5 mm

Upon attaching an auxillary mass m, to a main mass me through a spring of stiffness kz, the system looks like this: I to coswt m = 5kg m E = 57N . Trilt The enclosed system containing spring of stiffness K and mass m₂ is the dynamic vibration Absorber W = 66 rad/s K & K & ( 4 ) 7 ml I 2 Lt) The equation of motion of the masses in, and ma : må fk, x, & Kz (2-x) = F wsuit ? l m ärk, (x,-x) = 0 By assuming a harmonic solution X = X Coswt (s = 1, 2) where X is the steady state amplitude of my for S= 1, 2 , we use equations ① and 6 to get X, and X

X = (kz - m, ²) to (Kitky - mw*)(x,- m, w')-k . XzKxfeq tkitty mwa) eks-mW") experiance no • Since we want vibration at the ventilator to all, we want - X, 0 = K₂-m,w² =0 (from ③ ) using (16 in equation ④ The riegative sign indicates that kq X z opposes F. the spring force

the mass m₂ tonn We are also required to keep from braving an amplitude of - 1x, 1 = 6 = 10 .. lomm = m 10 1000 K 1000 =) K : 100 F = 100x57= 5.7 kN/m substituting k: 5700 N/m in (10 m2 = k = 5700 = 1.31 kg. Theis m₂ = 1.31 kg K = 5.7 kN/m (mass of the dynamic vibration absorber (stiffness of the dynamic vil absorber