The shock absorber of a truck is designed of a spring in parallel with a dashpot viscous damper. The truck is 20 tonnes and has a damping ratio of 125 and the spring stiffiess is 2 x 10" N/m. what is the damping coefficient? If the shock absorber is used

To find the damping coefficient \( c \) of the shock absorber, we can use the relationship between the damping ratio \( \zeta \), the mass \( m \), the spring stiffness \( k \), and the damping coefficient \( c \). The formula for the damping ratio is given by: \[ \zeta = \frac{c}{2\sqrt{mk}} \] Where: - \( m \) is the mass (in kg), - \( k \) is the spring stiffness (in N/m), - \( c \) is the damping coefficient (in Ns/m), - \( \zeta \) is the damping ratio. Given: - \( m = 20,000 \, \text{kg} \) - \( \zeta = 1.25 \) - \( k = 2 \times 10^5 \, \text{N/m} \) We can rearrange the equation to solve for \( c \): \[ c = 2\zeta\sqrt{mk} \] Substituting the known values: \[ c = 2 \cdot 1.25 \cdot \sqrt{20,000 \cdot 2 \times 10^5} \] First, calculate \( mk \): \[ mk = 20,000 \cdot 2 \times 10^5 = 4 \times 10^{10} \] Now take the square root: \[ \sqrt{mk} = \sqrt{4 \times 10^{10}} = 2 \times 10^5 \] Now substitute back into the equation for \( c \): \[ c = 2 \cdot 1.25 \cdot 2 \times 10^5 = 5 \times 10^5 \, \text{N·s/m} \] Therefore, the damping coefficient \( c \) is: \[ c = 500,000 \, \text{N·s/m} \] --- Now for the second part of the question, we are considering the shock absorber for a 5-tonne car. First, we need to calculate the new natural frequency \( \omega_n \) and the new damping ratio \( \zeta' \). The natural frequency \( \omega_n \) for a system is given by: \[ \omega_n = \sqrt{\frac{k}{m}} \] For the 5-tonne car: - Mass \( m' = 5 \, \text{tonne} = 5,000 \, \text{kg} \) - Spring stiffness \( k = 2 \times 10^5 \, \text{N/m} \) Calculating the natural frequency: \[ \omega_n = \sqrt{\frac{2 \times 10^5}{5,000}} = \sqrt{40} \approx 6.32 \, \text{rad/s} \] Now, calculate the new damping ratio \( \zeta' \) using the previously calculated damping coefficient \( c \): \[ \zeta' = \frac{c}{2\sqrt{m'k}} \] Substituting values: \[ \zeta' = \frac{500,000}{2 \sqrt{5,000 \cdot 2 \times 10^5}} \] Calculating \( m'k \): \[ m'k = 5,000 \cdot 2 \times 10^5 = 10^{9} \] \[ \sqrt{m'k} = \sqrt{10^{9}} = 10^{4.5} = 31622.78 \] Now substituting back to find \( \zeta' \): \[ \zeta' = \frac{500,000}{2 \cdot 31622.78} \approx \frac{500,000}{63245.56} \approx 7.91 \] Thus, the values are: - Natural Frequency \( \omega_n \approx 6.32 \, \text{rad/s} \) - Damping Ratio \( \zeta' \approx 7.91 \) In summary: - Damping coefficient \( c = 500,000 \, \text{N·s/m} \) - Natural frequency for the 5-tonne car \( \omega_n \approx 6.32 \, \text{rad/s} \) - Damping ratio for the 5-tonne car \( \zeta' \approx 7.91 \)