**1 Answer**

written 4.3 years ago by |

**The observed amplitudes on the same side of the neutral axis for successive cycles are found to decay 50% of the initial value.**

Determine:

1] Logarithmic decrement.

2] Damping torque per unit velocity.

3] The periodic time of vibration.

4] The frequency when the disc is removed from the fluid.

Let, initial displacement is $\theta_1 = \theta$

$\therefore$ $\delta = \frac{1}{n} \ ln \ (\frac{\theta}{0.5 \theta}) = \frac{1}{1} \ ln \ (\frac{1}{0.5}) = \ ln \ (2)$

$\delta = 0.693$

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

$\therefore$ $\xi = 0.109$

$\therefore$ $\xi = \frac{c}{\xi} = \frac{c}{2.I. w_n} = \frac{c}{2J \sqrt{ \frac{kt}{j}}}$

$\therefore$ [ NOTE: $k_t = \frac{GI}{L} = \frac{4.5 \times 10^{10} \times \frac{\pi }{32} (0.1)^4}{0.50} = 8.831 \times 10^5 Nm/rad$]

$\therefore$ $0.log = \frac{c}{2 \times 0.05 \times \sqrt{ \frac{8.831 \times 10^5}{0.05}}}$

C = 45.809 N.m s/rad

$\rightarrow$ $Td = \frac{2 \pi}{wd} = \frac{2 \pi}{w_n \sqrt{1 – \xi ^2}} = \frac{2\pi}{\sqrt{\frac{kt}{J}} \sqrt{1 – \xi^2}}$

$Td = \frac{2 \pi}{\sqrt{ \frac{8.831 \times 10^5}{0.05}}} \sqrt{ 1 – (0.109)^2}= 1.503 \times 10^{-3} sec$

$\rightarrow$ When the disc is removed from viscous fluid.

$w_n = 2\pi f_n = \sqrt{ \frac{kt}{J}} = \sqrt{ \frac{8.831 \times 10^5}{0.05}} = 4202.67 rad/sec$

$f_n = \frac{4202.67}{2 \pi } = 669.2 Hz$