Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is = [tex]5.676*10^{-3} \ mm[/tex]
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 [tex]\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}][/tex]
where;
[tex]\epsilon = 0[/tex]
[tex]W_n = \sqrt { \frac{k}{m}}[/tex]
= [tex]\sqrt { \frac{100*32.2}{200}}[/tex]
= 4.0124
replacing them into the above equation and making X the subject of the formula:
[tex]X =[/tex] [tex]Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}[/tex]
[tex]X =[/tex] [tex]0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}[/tex]
[tex]X =[/tex] [tex]5.676*10^{-3} \ mm[/tex]
Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = [tex]5.676*10^{-3} \ mm[/tex]
