Question

0.2 m = 2Y k 35 m

An automobile is considered as a system consisting of a mass-spring of single degree of freedom and dampener that makes vibration movement in the vertical direction. The car is driven on a road that its surface is sinusoidally changing as shown in the figure. The distance between the highest point and the lowest point on the road surface is 0.2 m, and the distance between two consecutive peaks along the road is 35 m. If the natural frequency of the automobile is 2 Hz and the damping ratio of the shock dampeners is 0.15, (a) Determine the vibration amplitude of the automobile at a speed of 60 km/h. (b) When the speed of the automobile is changed, find the most unfavorable (maximum amplitude vibration) speed value for the passengers.

Answer 1

Trào. " x= 35 m Given data: 0 Amplitude , Y = 0.1 m wavelength ,x=35m (3) Natural freguency, fo = 2 Hz (4) Damping ratio , {=0.15 solation O Vibration amplitude (X) at V = 60 km hr - 60 km hr Speed, V = boxs / V = 16.67 mong [ 1 leto - Foto ] 2 frequency of vibration, : fo 16.67 35 X f 0.476 Hz : frequency ratio , r = 0.476 f fr 2 r = 0.238 </ 14 (267) (1-72)2 +12&r)?

1+ (2 X0-15% 0.238)? xlo (1-0-2382) + (2x0.15X0.238)? :: Х 0:10597 m Ş0:10597 X:100 Х = 10.597 cm [Im Floo com] ® condition Most unfavourable speed value for passengers (Vc) When $ = Tь һәм ал critical and the speed which critical we get is called speed (vo. At this speed, amplitude of vibration becomes manimum and this is most unfavourable speed. at critical sked, f = fo fn - Vc ve .: 2 Vc. 35 . Ve= 70 m 5