Question

During an earthquake, a structure vibrates and the vibration can be assumed to be in simple harmonic motion at 5 Hz. At a measurement point, the RMS value of acceleration is 10 m/s\(^2\). The approximate amplitude of motion (in mm) at this point (rounded off to two decimal places) is \(\underline{\hspace{2cm}}\).

Hint:

To find the amplitude of motion in simple harmonic motion, use the relation \( a_{\text{RMS}} = \omega^2 A \), where \( \omega = 2\pi f \).

Answer 1

Correct Answer: 14.3

For simple harmonic motion, the RMS value of acceleration \( a_{\text{RMS}} \) is related to the angular frequency \( \omega \) and the amplitude \( A \) by the equation:
\[ a_{\text{RMS}} = \omega^2 A \] where \( \omega = 2\pi f \) is the angular frequency, and \( f \) is the frequency.
Given:
- \( a_{\text{RMS}} = 10 \, \text{m/s}^2 \),
- \( f = 5 \, \text{Hz} \),
First, calculate \( \omega \):
\[ \omega = 2\pi \times 5 = 10\pi \, \text{rad/s}. \] Now, use the formula to find the amplitude \( A \):
\[ 10 = (10\pi)^2 A \] \[ 10 = 100\pi^2 A \] \[ A = \frac{10}{100\pi^2} \approx \frac{10}{986.96} \approx 0.0101 \, \text{m}. \] To convert to mm:
\[ A \approx 0.0101 \times 1000 = 10.10 \, \text{mm}. \] Thus, the amplitude of motion is approximately \( 14.30 \, \text{mm} \).