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Department of Mechanical Engineering Technology and Industrial Design MECH 4760 Vibration & Advanced Dynamics Quiz 2 Name: Yue cuan, 204 Fall 2019 A vibrating system consisting of a mass of 5.4 kg is viscously damped such that the ratio of any two consecutive amplitudes is 0.98mm and 0.68mm. The frequency of the oscillation is found to be 9 Hz. (a) Determine the damping coefficient. (b) What is the amplitude after one cycle? (A = 12 mm and d = 0)

Answer 1

solution:- given, m= 5:4kg. viscoresly damped system x. 0.98 mm a frequency of damped. Oscillation, f = 9Hz. Wa = daniped oscillation frequency wd=zxf = 2xaxa = 56.54866 rad (see (a) since, the toitude of the in formula for logarithenic decrement amplitude of oscillation is given by een ( ) PEITO or, lu (og) = 27 1 2 o of 0.365450 JT p2 = 224 0.133 560846 (1-2) = 474g?

of 0.133.560846 = 39,61197845.? os oss 0.058066588 r ļ = 0.0561 Again os we = wnri-q2 56.54866 := win VI- 0.05812 or w 56.54866 VI-0.05823 = 56.64435 rad Isee. damping Hence formula for calculation of Coefficient, c = 2mqwn 245:48 0.0581 X 56.64435 = 35.54d. See - 35.5431. Nos .m .

sorte by Amplitude. After one cycle - 23 Now since, ln (12) = 243 24 og lan ( 9 ) = 2*** 0.05ELL TT-0.05572 os lui on = 0.345670729 os 68 mint = 7.44148 O = 47.1737 mm Hence, damping coefficient, c = 35.5431 Nisa Amplitude offer . X = 47. 1737 mm.... i and one cycles 13