Answer:
The amplitude is 0.010 m
Solution:
As per the question:
Eccentric Mass, [tex]m_{e} = 10\ kg[/tex]
[tex]m_{e}[/tex] = 10%M kg
0.1M = 10
Total mass, M = 100 kg
Spring constant, k = 3200 N/m
Eccentric center, e = 100 mm = 0.1 m
Speed of motor, N = 1750 rpm
Distance between two springs, d = 250 mm = 0.25 m
Now,
Angular velocity, [tex]\omega = \frac{2\pi N}{60} = \frac{2\pi \times 1750}{60} = 183.26\ rad/s[/tex]
For the vertical vibrations:
[tex]\omega_{n} = \sqrt{\frac{2k}{M}} = \sqrt{\frac{6400}{100}} = 8\ rad/s[/tex]
Now, the frequency ratio is given by:
r = [tex]\frac{\omega}{\omega_{n}} = \frac{183.26}{8} = 22.91[/tex]
Thus the amplitude is given by:
A = [tex]e^{\frac{m_{e}}{M}}.\frac{r^{2}}{|1 - r^{2}|}[/tex]
A = [tex]e^{\frac{10}{100}}.\frac{22.91^{2}}{|1 - 22.91^{2}|} = 0.010 m[/tex]
