The amplitude of natural vibration decreases to one fourth of the initial value after 5 oscillations,

Determine:

1] The logarithmic decrement.

2] The damping force and damping co-efficient.

3] The stiffness of the spring.

$x_6 = \frac{1}{4} x_1$

$\therefore$ $\frac{x_1}{x_6} = 4$

$\rightarrow$ $fd = \frac{No \ of \ Oscillations}{Time} = \frac{45}{27} = 1.67 Hz.$

$wd \ = \ 2\pi \ fd \ = \ 2\pi \ \times \ 1.67 \ = \ 10.471 $ rad/s

$\rightarrow$ Logarithmic decrement.

$\delta \ = \ \frac{1}{n} \ ln \ (\frac{x_1}{x_c})$

$\frac{1}{5} \ ln \ (4)$

$\delta \ = \ 0.277$

$\rightarrow$ damping force and damping co-efficient.

$\delta = \frac{2 \pi \xi}{ \sqrt{1 – \xi^z}}$

$0.277 = \frac{2 \pi \xi}{\sqrt{1 = \xi^2}}$

$\xi = 0.044$

$w_d = w_n \sqrt{1 – \xi^2}$

$\frac{10.47}{\sqrt{1- (0.044)^2}} = w_n$

$w_n = 10.48$ rad/sec.

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

$10.48 = \sqrt{ \frac{k}{16}}$

K = 1757.28 N/m

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

$C = \xi \times 2m \ w_n$

$= 0.044 \times 2 \times 16 \times 10.48$

$c = 14.75 \frac{NS}{m}$

$c = 14.75 \frac{N}{m/sec}$

C = 14.75 N

Damping force per unit velocity = F = 14.75N