An automobile is modeled as a singledegreeoffreedom system f

An automobile is modeled as a single-degree-of-freedom system free to vibrate in the vertical direction. In a free-vibration test, the amplitude of oscillation is observed to decrease from 20mm to 2mm in one cycle which takes 1/3s. When carrying passengers with a total mass 300kg, the free-vibration amplitude is observed to decay from 20mm to 2.75mm in one cycle. Find the mass of the vehicle and the stiffness and the damping of the suspension system

Solution

Solution :-

Given

T₁ =1/3 second =0.33 second

Passengers total mass = 300 kg

Let mass of vehicle = M kg

Let stiffness = K

T₁ = 2 (M/K)

0.33 = 2(M/K)                     . . . .. (1)

T₂ = 2 ((M+300)/K)              ..   .. … (2)

We know that

k(number of cycle) * § = ln(20/2.75)

1* § = ln(20/2.75)

§ = 1.98

We know that

§ = 2 p

p = Damping ratio

p =1.98/(2) = 0.315

it is less than 1 so system is under damped

we know that

T₂ = 2 /(_n(1- p²))                 ………..   (3)

_n   = 2/T₁

      = 2/(0.33) = 19.04 radian/second

T₂ = 2/(19.04(1- (0.315)²))

T₂ = 0.347 second

On putting value of T₂ = 0.347 seconds in equation (2)

0.347 = 2 ((M+300)/K)                  ..   .. ..(4)

From equation (1)

M/K = 2.75 *10^-3

M = 2.75 * 10^-3 K

On putting value of M in equation (4)

0.347 = 2 ((2.75*10^-3 *K + 300)/K)

Stiffness(K) = 10⁶ N/m   Answer

M = 2.75 *10 ^-3 *10⁶

( Mass of vehicle )M = 2750 Kg   Answer

Damping of suspension system

p = C/C_cr

Damping (C) = p * C_cr

                       = p * 2 (K*M)

                       = 0.315 * 2(10⁶ * 2750)

    Damping (C) = 33.03 * 10³ N-S/m     Answer