**1 Answer**

written 5.4 years ago by |

**The damped period of vibration is limited to 2 sec and the amplitude of vibration should reduce to one sixteen in one cycle,**

**Find:**

a] Spring stiffness.

b] Damping co efficient of the shock absorber.

$x_2 = \frac{1}{16} x_1$

n = 1

$x_2 = \frac{1}{16} x_1$

$\therefore$ $\frac{x_1}{x_2} = 16$

m = 200 kg

$T_d$ = 2 sec

$\delta = ln (\frac{x_1}{x_2}) = \ ln (16) = 2.77$

$\delta = \ \frac{2 \pi c_c}{\sqrt{1 - c_c^2}}$

$2.77 \ = \ \frac{2 \pi c_c}{\sqrt{1 - e_c^2}}$

$\xi$ = 0.4033

$w_n = \sqrt{ \frac{k}{m}}$

$w_d = w_n \sqrt{1 - \xi^2}$

$w_d = \frac{2 \pi }{T_d}$

$w_d = \frac{2 \pi }{2} = 3.14$ rad/s

$w_n = \frac{w_d}{\sqrt{1- \xi^2}}$

$= \frac{3.14}{\sqrt{1- 0.403^2}}$

$w_n = 3.43$ rad/sec

$w_n = \sqrt{\frac{k}{m}}$

$3.43 = \sqrt{\frac{k}{200}}$

K = 2352.98 N/m - - - Ans (1)

$\xi = \frac{c}{C_c} = \frac{c}{2m \ w_n}$

$0.403 = \frac{c}{2 \times 200 \times 3.43}$

C = 552.91 Ns/m - - - Ans (2)