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Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer: A seismic instrument has a natural frequency \(\mathrm{{30}{H}{z}}\)30Hz and a damping ratio \(\mathrm{{0.75}}\)0.75 , the amplitude error the indicated displacement less than \(\mathrm{{5}\%}\)5% then find the maximum permitted frequency if the device acts a accelerometer.

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Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer: A seismic instrument has a natural frequency \(\mathrm{{30}{H}{z}}\)30Hz and a damping ratio \(\mathrm{{0.75}}\)0.75 , the amplitude error the indicated displacement less than \(\mathrm{{5}\%}\)5% then find the maximum permitted frequency if the device acts a accelerometer.

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Asked by: manish kumar
Created at: 2025-10-12 16:17:45

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# Step-by-Step Solution Let's break down the problem: - **Natural frequency** (\(f_n\)) = 30 Hz - **Damping ratio** (\(\xi\)) = .75 - **Amplitude error** must be less than 5% (\(05\)), meaning: \[ \left| \{X}{X_} - 1 \right| < .05 \] where \(X\) is the indicated amplitude and \(X_\) is the true amplitude. - The device acts as an **accelerometer**, so we consider its high-frequency response. --- ## 1. **System Response as an Accelerometer** For a damped single degree-of-freedom system, the amplitude ratio when subjected to base motion (for accelerometer use) is: \[ \frac{X}{X_} = \frac{r^2}{\sqrt{(1 - r^2)^2 + (2\xi r)^2}} \] where: - \(r = \frac{f}{f_n}\) - \(f\) is the frequency of the input (ground motion) - \(f_n\) is the natural frequency --- ## 2. **Amplitude Error Condition** Since we want the **amplitude ratio to be within 5% of 1**, set up the inequality: \[ \left| \frac{X}{X_} - 1 \right| < .05 \] Let’s use the high-frequency approximation, since for an accelerometer, \(f \gg f_n\), so \(r \gg 1\): \[ \frac{X}{X_} \approx 1 - \frac{1}{2\xi^2 r^2} \] Set the error less than 5%: \[ \left| 1 - \frac{1}{2\xi^2 r^2} - 1 \right| < .05 \] \[ \frac{1}{2\xi^2 r^2} < .05 \] --- ## 3. **Solving for Maximum Permitted Frequency** Substitute \(\xi = .75\): \[ \frac{1}{2 \times (.75)^2 \times r^2} < .05 \] \[ 2 \times (.75)^2 = 2 \times .5625 = 1.125 \] \[ \frac{1}{1.125 \cdot r^2} < .05 \] \[ 1 < .05 \cdot 1.125 \cdot r^2 \] \[ 1 < .05625 \cdot r^2 \] \[ r^2 > \frac{1}{.05625} \] \[ r^2 > 17.78 \] \[ r > \sqrt{17.78} \approx 4.22 \] Recall \(r = \frac{f}{f_n}\), so: \[ \frac{f}{30} > 4.22 \] \[ f > 4.22 \times 30 = 126.6~\text{Hz} \] --- ## 4. **Final Answer** > **The maximum permitted frequency is approximately \(\boxed{126.6~\mathrm{Hz}}\) for the amplitude error to be less than 5%.** --- ### Summary Table | Parameter | Value | |------------------|-----------| | Natural frequency| 30 Hz | | Damping ratio | .75 | | Max permitted freq| 126.6 Hz | --- ## **How to Use This Result** - For frequencies **above 126.6 Hz**, the amplitude error will be less than 5% when the device is used as an accelerometer with these settings. - For lower frequencies, the error will exceed the 5% threshold. --- **Visual:** ![Accelerometer response curve, showing error band](https://upload.wikimedia.org/wikipedia/commons/7/73/Second_order_transfer_function_magnitude_response.png) *Alt text: Typical frequency response curve for a second-order system, with error threshold indicated.*

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