**1 Answer**

written 5.0 years ago by |

**The table is bolted to the laboratory floor. Measurement indicate that the floor has a steady state vibration amp of 0.5 mm at a frequency of 30 Hz. What is the amp of acc of the flow monitoring device?**

m = 35kg

k = $2 \times 10^5 N/m$

$\xi = 0.08$

Y = $0.5 \times 10^{-3} \ m$

f = 30 Hz

w = $2 \pi f = 2 \pi \times 30 = 188.49$ rad/s.

$\rightarrow$ Acceleration of flow monitoring device.

$= w^2. X$

$\frac{x}{y} = \frac{\sqrt{1+(2 \xi r)^2}}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

$w_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{ 2 \times 10^5}{35}} = 75.59$ rad/sec

$r = \frac{188.49}{75.59} = 2.49$

$\frac{x}{0.5 \times 10^{-3}} = \frac{\sqrt{ 1+ (2 \times 0.08 \times 2.49)^2}}{\sqrt{(102.492^2)^2 + (2 \times 0.08 \times 2.49)^2}}$

= $\frac{1.0766}{5.215}$

= 0.2064

$X = 1.03 \times 10^{-4} \ m$

Acceleration of flow monitoring device = $w^2.X$

$= (188.49)^2 \times (1.03 \times 10^{-4})$

$= 3.67 \ m/sec^2$

**Vibration Isolation and Transmissibility.**

1] $T_r \ = \ \frac{FT_r}{F_o} \ = \ mo.e.w^2$

$T_r \ = \ \frac{ \sqrt{1 + (2 \xi r)^2}}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

To find mo.e

$\frac{X}{\frac{mo.e}{m}} = \frac{r^2}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

2] $\eta_{iso} = 1 - Tr$

3] $tan \ \phi = \frac{2 \xi r}{1 – r^2}$ [phase angle]

4] $tan \ \alpha = \frac{cw}{k}$ [phase lag]

5] Angle between Fo and Frr

$\psi = \phi - \alpha$