**1 Answer**

written 5.3 years ago by |

**Damping is negligible and unit is constrained to move vertically.**

**Determine dynamic amplitude of the machine.**

**Redesign the mountings so that the dynamic amplitude is reduced to one half of the original value, but maintain same natural frequency.**

W = 1100N, $m = \frac{1100}{9.81} = 112.13$ kg

$m_o = \frac{220}{9.81} kg = 22.43 kg$

$e = 5 \times 10^{-3} \ m$

N = 600 rpm

$K = 88 \times 10^3$

$\xi = 0$

Now, $w = \frac{2 \pi N}{60} = \frac{2 \pi \times 600}{60} = 62.83$ rad/sec.

Find - $X_1$ = ?

Redesign mountings when new amplitude, $X_2 = \frac{x1}{2}$

**Case 1-**

$\frac{X_1}{\frac{mo.e}{m}} = \frac{r^2}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

$\frac{X1}{\frac{22.43 \times 5 \times 10^-3}{112.13}} = \frac{r^2}{1-r^2}$

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

$Wn = \sqrt{ \frac{88 \times 10^3}{112.13}}$

$Wn = 28.01$ rad/sec

$r = \frac{w}{wn} = \frac{62.83}{28.01}$

R = 2.242

Since r > 1

$\therefore$ RHS will be

$\frac{X1}{1 \times 10^{-3}} = \frac{r^2}{r^2 – 1}$

$\frac{X1}{1 \times 10^{-3}} = \frac{r^2}{r^2 – 1}$

$X_1 = 1.25 \times 10^{-3} \ m$

**NOTE:** To maintain same frequency (Wn) let us introduce a damper with damping ratio.

$\frac{6.23 \times 10^{-4}}{1 \times 10^{-3}} = \frac{(2.242)^2}{\sqrt{ (1- 2.242)^2 + ( 2 \times \xi \times 2.242)^2}}$

$0.623 = \frac{5.02}{\sqrt{16.24 + 4 \xi^2 \times 5.02}}$

$0.623 = \frac{5.02}{\sqrt{16.24 + 20.08 \xi^2}}$

$16.24 + 20.08 \xi^2 = 64.93$

$\xi = 1.55$