written 5.3 years ago by |
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