A mass of 10 kg is kept on two slabs of isolators placed one over
| A mass of 10 kg is kept on two slabs of isolators placed one over the other. One of the isolators is of rubber having a stiffness of 3 kN/m and damping coefficient of 100 N-s/m. While the other isolator is of felt with stiffness of 12 kN/m and damping coefficient of 300 N-s/m. If the system is in motion on vertical direction, find
A. The damping factor of the system 0.35
B. The damping factor of the system 0.24
C. The equivalent stiffness is 2400 N/m
D. The equivalent stiffness is 75 N/m
Please scroll down to see the correct answer and solution guide.
Right Answer is:
SOLUTION
Explanation:
The isolators are connected in series
\(\frac{1}{{{K_e}}} = \frac{1}{{3000}} + \frac{1}{{12000}}\)
∴ Ke = 2400 N/m
Now,
\(\frac{1}{{{C_e}}} = \frac{1}{{{c_1}}} + \frac{1}{{{c_2}}}\)
\(\frac{1}{{{C_e}}} = \frac{1}{{100}} + \frac{1}{{300}}\)
∴ Ce = 75 N-s/m
Now,
\(\xi = \frac{{{C_e}}}{{2\sqrt {{K_e}m} }} = \frac{{75}}{{2\sqrt {2400\; \times \; 10} }}\)
∴ ξ = 0.24