Question

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A 200 kg machine is mounted on springs whose stiffness is 5x105 N/m. The machine has an exciting force of 1000sin(53t). (a) What is the steady-state amplitude if no vibration absorber is added? (b) Design a vibration absorber of mass 40 kg such that the steady-state amplitude of the primary system is zero at 53 rad/s. (c) What are the natural frequencies of the two degree of freedom system when the absorber designed in part (b) is used? (d) With the absorber designed in part (b) in place what is the steady-state amplitude of the rimary system is the frequency of excitation is 50 rad/s. (e) Design a vibration absorber of minimum mass such that the steady-state amplitude of the absorber is 5 mm and vibrations of the primary system are eliminated.

Answer 1

Given that m = 200 kg k = b5 times 10^5 N/m F = = 100 sin (53 t) tilled = ncos un (omega T) a) steady state amplitude if no vibration absorver is added , c = 0 amplitude = f_o/k /square ropot (1-r^2) + (2cr) ^2 where T = c/2 squareroot c/2 km = damping ratio 18 c = 0 t = 0 amplitude = fo/k (1-r^2) r = frequencuy ratio = w/wn r = frequency ratio = w /wn wn = squareroot k/m = squareroot 5 times 10^5/200 = 50 rad/ sec w = 53 (given) r = w/wn = 53/50 = 1.06 amplitude = fo/k/r^2 -1 = 1000/(5 times 10^5) /(1-00-1) x = = 16.18 mm (b) if the steady state amplitude of primary system is max then for max amplitude condition will be x_st = fo/ k/ squareroot (1-r^2) + (2yr) ^2 dx/dtheta = 0 2 (1 r^2 times 2r 2c times2c times 2r = 0 r = squareroot 1- 25 ^2 greater than 0) t less than or equalt o 1/ squareroot 2 = t = 0.707