Question

In order to cut down on vibratory motion transmitted to instruments in aircraft, the instruments are often mounted on isolators. Given a case where an isolator, having negligible damping, deflects 1/8 inches under the weight of a given instrument of 50 lbs, find percentage of motion transmitted to that instrument in the vibration of aircraft is at a frequency of 2000 rpm?

Answer 1

S = 0.125 in = 3.175 times 10^-3 m W = 50 lbf N = 2000 rpm omega_n = squareroot g/s = squareroot 9.81/3.175 times 10^-3 omega_n = 55.6 rad/s w = 2 pi N/60 = 2 pi times 2000/60. Omega = 209.44 rad/s. R = omega/ omega_n = 209.44/55.6 r = 3.766 F_T/F_0 = squareroot 1 + (2yr)^2 /squareroot (1 - r^2)^2 + (2 yr)^2 F_T/F_0 = 1/ plusminus (1 - r^2) F_T/F_0 = 1/3.766^2 - 1 F_T/F_0 = 0.0758 implies F_T/F_0 % = 7.6 % For maximum vibration transmission - r = squareroot 1 + (2 y)^2 i = 1 r = 1 implies omega = omega_n 2 pi N/60 = 55.6 implies N = 530.24 spm