Question

A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibration amplitude of 0.05mm at 2 Hz (a typical resonant frequency of a building). a) If there is no damping, how much transmitted vibration (amplitude of displacement) does the microscope experience

Answer 1

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = $5.676*10^{-3} \ mm$

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  $\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]$

where;

$\epsilon = 0$

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

= $\sqrt { \frac{100*32.2}{200}}$

= 4.0124

replacing them into the above equation and making X the subject of the formula:

$X =$ $Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}$

$X =$ $0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}$

$X =$ $5.676*10^{-3} \ mm$

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = $5.676*10^{-3} \ mm$