Question
What is the maximum stiffness of an isolator of damping ratio 0.1 such that the acceleration felt by the apparatus of Problem 8.19 is less than $6 \mathrm{~m} / \mathrm{s}^2$ ?
Step 1
The formula is: \[ k_{\text{max}} = \frac{4m\zeta\omega_n^2}{(1-2\zeta^2)} \] where \( k_{\text{max}} \) = maximum stiffness \( m \) = mass of the apparatus \( \zeta \) = damping ratio \( \omega_n \) = natural frequency of the system Show more…
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Applications
A sensitive instrument of mass $100 \mathrm{~kg}$ is installed at a location that is subjected to harmonic motion with frequency $20 \mathrm{~Hz}$ and acceleration $0.5 \mathrm{~m} / \mathrm{s}^{2}$. If the instrument is supported on an isolator having a stiffness $k=25 \times 10^{4} \mathrm{~N} / \mathrm{m}$ and a damping ratio $\zeta=0.05,$ determine the maximum acceleration experienced by the instrument.
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