Question:medium

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

Show Hint

Compare with \( \ddot{X} + \omega^2 X = 0 \), then:
\( f = \dfrac{\omega}{2\pi} \)
Updated On: Mar 20, 2026
  • \(3 \, \text{Hz}\)
  • \(6 \, \text{Hz}\)
  • \(12 \, \text{Hz}\)
  • \(18 \, \text{Hz}\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
We are given the differential equation for an undamped free vibration system and asked to calculate its natural frequency (\(f\)) in Hertz (Hz).
Step 2: Key Formula or Approach:
The standard equation of motion for a simple harmonic oscillator (free vibration system) is:
\[ \ddot{X} + \omega_n^2 X = 0 \] where \(\omega_n\) is the natural angular frequency in radians per second (rad/s).
The natural frequency \(f\) in Hertz is related to \(\omega_n\) by:
\[ f = \frac{\omega_n}{2\pi} \] Step 3: Detailed Explanation:
We compare the given equation with the standard form:
Given equation: \( \ddot{X} + 36\pi^2 X = 0 \)
Standard form: \( \ddot{X} + \omega_n^2 X = 0 \)
By comparing the coefficients of \(X\), we get:
\[ \omega_n^2 = 36\pi^2 \] Taking the square root of both sides gives the natural angular frequency:
\[ \omega_n = \sqrt{36\pi^2} = 6\pi \, \text{rad/s} \] Now, we convert the angular frequency \(\omega_n\) to natural frequency \(f\):
\[ f = \frac{\omega_n}{2\pi} = \frac{6\pi}{2\pi} = 3 \, \text{Hz} \] Note: The provided answer key in the source code seems to indicate 6 Hz, but the calculation correctly yields 3 Hz, which corresponds to option (A). We will proceed with the mathematically correct answer. Step 4: Final Answer:
The natural frequency of the system is 3 Hz. Therefore, option (A) is the correct answer.
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