Question

In a tractor seat system, the chassis frequency and seat suspension damping rate are 20 rad s\(^{-1}\) and 400 N m\(^{-1}\) s, respectively. The critical damping rate of the tractor seat system is 1600 N m\(^{-1}\) s. If the combined mass of the seat and operator is 80 kg, the transmissibility of vibration is \underline{\hspace{2cm}} (rounded off to 2 decimal places).

Hint:

Always calculate natural frequency, damping ratio, and frequency ratio step by step. Substitution into the transmissibility formula gives precise results.

Answer 1

Step 1: Natural frequency. The stiffness of the system is obtained from the critical damping coefficient: \[ c_{cr} = 2m\omega_n \] \[ 1600 = 2 \cdot 80 \cdot \omega_n \Rightarrow \omega_n = 10 \, rad/s \]

Step 2: Frequency ratio. \[ r = \frac{\omega}{\omega_n} = \frac{20}{10} = 2 \]

Step 3: Damping ratio. \[ \zeta = \frac{c}{c_{cr}} = \frac{400}{1600} = 0.25 \]

Step 4: Transmissibility formula. \[ T = \frac{\sqrt{1+(2\zeta r)^2}}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}} \] Substitute: \[ T = \frac{\sqrt{1 + (2 \cdot 0.25 \cdot 2)^2}}{\sqrt{(1-2^2)^2 + (2 \cdot 0.25 \cdot 2)^2}} \] \[ T = \frac{\sqrt{1+1^2}}{\sqrt{9+1}} = \frac{\sqrt{2}}{\sqrt{10}} = 0.447 \] \[ \boxed{0.45} \]