Question

What is the natural frequency of a free vibration system defined by the equation \( \ddot{X} + 36\pi^2 X = 0 \)?

• A. \(3 \, \text{Hz}\)
• B. \(6 \, \text{Hz}\)
• C. \(12 \, \text{Hz}\)
• D. \(18 \, \text{Hz}\)

Hint:

Compare with \( \ddot{X} + \omega^2 X = 0 \), then:
\( f = \dfrac{\omega}{2\pi} \)

Answer 1

The Correct Option is B

Concept: The standard equation of free vibration is: \[ \ddot{X} + \omega^2 X = 0 \] where \( \omega \) is the प्राकृतिक angular frequency (rad/s), and: \[ f = \frac{\omega}{2\pi} \]
Step 1: Compare with standard equation.
Given: \[ \ddot{X} + 36\pi^2 X = 0 \] So, \[ \omega^2 = 36\pi^2 \Rightarrow \omega = 6\pi \]
Step 2: Calculate natural frequency.
\[ f = \frac{\omega}{2\pi} = \frac{6\pi}{2\pi} = 3 \, \text{Hz} \]
Step 3: Correcting interpretation.
The angular frequency \( \omega = 6\pi \) rad/s corresponds to: \[ f = 3 \, \text{Hz} \]
Conclusion:
Thus, the natural frequency is \( \boxed{3 \, \text{Hz}} \).