Question 1.7.3: As a final example, consider modeling the vertical suspensio......

The mass is 1361 kg and the natural frequency is

so that

k = 1361 w²_{n}

At rest, the car’s springs are compressed an amount Δ, called the static deflection, by the weight of the car. Hence, from a force balance at static equilibrium, m\mathtt{g} = kΔ, so that

k = \frac{m\mathtt{g}}{\Delta }

and

w_{n}  =  \sqrt{\frac{k}{m}}  =  \sqrt{\frac{\mathtt{g}}{\Delta }}  =  \sqrt{\frac{9.8}{0.05}} = 14 rad/s

The stiffness of the suspension system is thus

k = 1361(14)² = 2.668 × 10^{5} N/m

For ζ = 0.3, equation (1.30) becomes

ζ = \frac{c}{c_{cr}}  =  \frac{c}{2mw_{n}}  =  \frac{c}{2\sqrt{km}}                  (1.30)

c = 2ζmw_{n} = 2(0.3)(1361)(14) = 1.143 × 10^{4} kg/s

Now if the passengers and luggage are added to the car, the mass increases to 1361 + 290 = 1651 kg. Since the stiffness and damping coefficient remain the same, the new static deflection becomes

Δ = \frac{m \mathtt{g}}{k} = \frac{1651(9.8)}{2.668  ×  10^{5}} ≈ 0.06 m

The new frequency becomes

w_{n}  =   \sqrt{\frac{\mathtt{g}}{\Delta }}  =  \sqrt{\frac{9.8}{0.06}} = 12.78 rad/s

From equations (1.29) and (1.30), the damping ratio becomes

c_{cr}  =  2mw_{n}  =  2\sqrt{km}         (1.29)

ζ = \frac{c}{c_{cr}}  =  \frac{c}{2mw_{n}}  =  \frac{c}{2\sqrt{km}}                  (1.30)

\zeta  =  \frac{c}{c_{cr}}  =  \frac{1.143  ×  10^{4}}{2mw_{n}}  =  \frac{1.143  ×  10^{4}}{2(1651)(12.78)} = 0.27

Thus the car with passengers, fuel, and luggage will exhibit less damping and hence larger amplitude vibrations in the vertical direction. The vibrations will take a little longer to die out.