AER 403 Lecture Notes - Lecture 18: Vibration Isolation, Interval Ratio, Inverse Trigonometric Functions

• School: Ryerson University
• Department: Aerospace
• Course: AER 403
• Professor: Fengfeng Xi

Document Summary

Mechanisms & vibrations: forced vibrations, support motion, vibration isolation vibration sensors. Course outline: support motion, vibration isolation vibration sensors. In this case, the excitation is a vibrating base. Examples include: ground vibration, aircraft body vibration car on a bumpy road, ship on waves, earthquake. The equation of motion is md2x/dt2 = -k(x-y) c(dx/dt dy/dt) Since the base is vibrating y=ysin(cid:600)t d2y/dt2 = -m(cid:600)(cid:884)ysin(cid:600)t md(cid:884)z/dt(cid:884) + cdz/dt + kz = m(cid:600)(cid:884)ysin(cid:600)t. This equation is similar to the equation of motion for imbalance induced vibration. Now the solution of z is converted to x. Based on the law of cosines, then x amplitude is x(cid:884) = z(cid:884)+y(cid:884)+(cid:884)zycos . = [(cid:883)+(cid:523)(cid:884) (cid:600)/(cid:600))2]/[(1 (cid:523)(cid:600)/(cid:600))2)2+(2 (cid:600)/(cid:600))2] tan(cid:599)=zsin /(cid:523)zcos +y) sin =(cid:523)c(cid:600)(cid:524)/[(cid:523)k-m(cid:600)(cid:884)(cid:524)(cid:884)+(cid:523)c(cid:600)(cid:524)(cid:884)](cid:883)/(cid:884); cos =(cid:523)k-m(cid:600)(cid:884)(cid:524)/[(cid:523)k- m(cid:600)(cid:884)(cid:524)(cid:884)+(cid:523)c(cid:600)(cid:524)(cid:884)](cid:883)/(cid:884) tan(cid:599)=zsin /(cid:523)y+zcos (cid:524) = mc(cid:600)(cid:885)/[(cid:523)k(cid:884)-km(cid:600)(cid:884)+(cid:523)c(cid:600)(cid:524)(cid:884)] x=xsin(cid:523)(cid:600)t-(cid:599)(cid:524) Original equation md(cid:884)x/dt(cid:884) + cdx/dt + kx = ky + cdy/dt y=ysin(cid:600)t, dy/dt = y(cid:600) cos(cid:600)t md(cid:884)x/dt(cid:884) + cdx/dt + kx = ky = kysin(cid:600)t x(cid:883)=(cid:523)f/k(cid:524)/