# 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 \( \zeta' \approx 7.91 \)