# Step-by-Step Solution Let's break down the problem: - **Given:** - Natural frequency, \( f_n = 30\,\text{Hz} \) - Damping ratio, \( \z = .75 \) - Allowable amplitude error in indicated displacement, \( < 5\% \) - **Asked:** Determine the **maximum input frequency** so that the amplitude error is less than 5%. --- ## 1. **Amplitude Ratio for a Damped System** For a seismic instrument (mass-spring-damper system), the amplitude ratio (also called **dynamic magnification factor**) is: \[ \left| \frac{X}{X_} \right| = \frac{1\sqrt{ [1 - r^2]^2 + [2\zeta r]^2 }} \] where: - \( X \): amplitude of the instrument's reading - \( X_ \): actual amplitude of ground motion - \( r = \frac{f}{f_n} \): frequency ratio (\(f\) is input frequency, \(f_n\) is natural frequency) - \( \zeta \): damping ratio --- ## 2. **Amplitude Error Condition** The **amplitude error** is: \[ \text{Amplitude error} = \left| 1 - \frac{X}{X_} \right| \] We require: \[ \left| 1 - \left| \frac{X}{X_} \right| \right| < .05 \] --- ## 3. **Set Up the Inequality** Let \( A = \left| \frac{X}{X_} \right| \): \[ |1 - A| < .05 \implies .95 < A < 1.05 \] But for \( f < f_n \) (which is the case for a seismic instrument), \( A < 1 \), so: \[ A > .95 \] --- ## 4. **Express \( A \) in Terms of \( r \) and \( \zeta \)** \[ A = \frac{1}{\sqrt{ [1 - r^2]^2 + [2\zeta r]^2 }} \] So, for \( A > .95 \): \[ \frac{1}{\sqrt{ [1 - r^2]^2 + [2\zeta r]^2 }} > .95 \] Take reciprocal and square both sides: \[ \sqrt{ [1 - r^2]^2 + [2\zeta r]^2 } < \frac{1}{.95} \] \[ [1 - r^2]^2 + [2\zeta r]^2 < \left( \frac{1}{.95} \right)^2 \] \[ [1 - r^2]^2 + [2\zeta r]^2 < 1.108 \] --- ## 5. **Plug in \(\zeta = .75\)** \[ [1 - r^2]^2 + [2 \times .75 \times r]^2 < 1.108 \] \[ [1 - r^2]^2 + [1.5r]^2 < 1.108 \] \[ [1 - r^2]^2 + 2.25 r^2 < 1.108 \] Let \( x = r^2 \): \[ (1 - x)^2 + 2.25 x < 1.108 \] \[ 1 - 2x + x^2 + 2.25x < 1.108 \] \[ x^2 + .25x + 1 - 1.108 < \] \[ x^2 + .25x - .108 < \] --- ## 6. **Solve the Quadratic Inequality** \[ x^2 + .25x - .108 = \] Use quadratic formula: \[ x = \frac{-.25 \pm \sqrt{(.25)^2 4 \times 1 \times (-.108)}}{2} \] \[ x = \frac.25 \pm \sqrt{.0625 + .432}}{2} \] \[ x = \frac{-.25 \pm \sqrt{.4945}}{2} \] \[ \sqrt{.4945} \approx .703 \] So, \[ x_1 = \frac{-.25 + .703}{2} \approx \frac{.453}{2} = .2265 \] \[ x_2 = \frac{-.25 - .703}{2} \approx \frac{-.953}{2} = -.4765 \] Only positive root is meaningful (since \( x = r^2 \ge \)). --- ## 7. **Find Maximum Frequency Ratio** \[ r^2 < .2265 \implies r < \sqrt{.2265} \approx .476 \] --- ## 8. **Calculate Maximum Input Frequency** Recall: \[ r = \frac{f}{f_n} \] \[ f < .476 \times f_n \] \[ f < .476 \times 30\,\text{Hz} = 14.28\,\text{Hz} \] --- # **Final Answer** ## The maximum input frequency for the vibration, such that the amplitude error in the indicated displacement is less than 5%, is: \[ \boxed{14.3~\text{Hz}} \] (Rounded to one decimal place.) --- ### **Summary Table** | Parameter | Value | |------------------------|---------------------| | Natural Frequency (\(f_n\)) | 30 Hz | | Damping Ratio (\(\zeta\)) | .75 | | Max Input Frequency (\(f\)) | \(\approx 14.3\) Hz| --- **If you want to see a plot of amplitude ratio vs frequency ratio, let me know!**