When the pendulum is vibrating, the observed amplitudes on the same side of the rest position for successive cycles are 9˚, 6˚ and 4˚.

Find:

1] Logarithmic decrement.

2] Damping torque at unit velocity.

3] Periodic time of vibration. Assume the modulus of rigidity $4.4 \times 10^{10} \ N/m^2$

What would the frequency be if the disc is removed from viscous fluid.

I = 0.06 $kgm^2$

L = 0.4

N = 2

$\theta_1$ = 9 ˚

$\theta_2$ = 6 ˚

$\theta_3$ = 4 ˚

[ d = 0.1 m

$G = 4.4 \times 10^{10} \ N/m^2$]

$\delta = \frac{1}{1} \ ln \ (\frac{\theta_1}{\theta_2}) \ = \ ln \ (\frac{9}{6})$

$\delta = 0.4054$

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

$0.4054 = \frac{2 \pi \xi}{\sqrt{ 1 – \xi^2}}$

$\xi = 0.064$

$wd = w_n \times \sqrt{ 1 – \xi ^2} = 4233.78 $ rad/sec

$kt = \frac{T}{\theta} = \frac{GJ}{L} = \frac{4.4 \times 10^{10} \times \frac{\pi}{32} \times (0.1)^4}{0.4}$

$Kt = 1.07 \times 10^6 \ Nm / rad$

$wn = \sqrt{ \frac{kt}{I}} = \sqrt{ \frac{1.07 \times 10^6}{0.06}} = 4242.48$ rad/sec

$\xi = \frac{Cr}{CcT}$

$0.064 \times 2 \times 0.06 \times 4242.8 = c_1$

cT = 32.43 Nm.sec/rad

Damping torque unit velocity = 32.43 Nm

$T_d = \frac{2 \pi}{wd} = \frac{2 \pi }{4233.78}$

$\frac{T_d = 1.48 \times 10^{-3} sec}{fn = \frac{wn}{2 \pi}}$

= $\frac{4242.48}{2 \pi }$

= 675.21 Hz.