Damped Vibration Equation Basics

The standard form of the equation of motion for a damped vibration system is a second-order linear differential equation:

$$ m\ddot x + c\dot x + kx = 0 $$

In this equation:

• m is the mass.
• c is the damping coefficient.
• k is the stiffness.
• x represents displacement, \(\dot x\) represents velocity, and \(\ddot x\) represents acceleration.

To analyze the damping characteristics, we often rewrite the equation in a normalized form by dividing by the mass m:

$$ \ddot x + \frac{c}{m}\dot x + \frac{k}{m}x = 0 $$

This normalized equation is then compared to the standard form representing damped oscillations:

$$ \ddot x + 2\zeta\omega_n\dot x + \omega_n^2 x = 0 $$

where:

• \(\omega_n = \sqrt{k/m}\) is the natural frequency of the undamped system.
• \(\zeta\) is the damping ratio, also known as the damping factor. It is a dimensionless parameter indicating how oscillations decay.

Given Equation Analysis

The specific equation provided in the question is:

$$ 6\ddot x + 9\dot x + 27x = 0 $$

To compare this equation with the standard forms, we first normalize it by dividing all terms by the coefficient of the acceleration term (\(\ddot x\)), which is the mass m = 6:

$$ \frac{6\ddot x}{6} + \frac{9\dot x}{6} + \frac{27x}{6} = 0 $$

This simplifies the equation to:

$$ \ddot x + 1.5\dot x + 4.5x = 0 $$

Calculating the Damping Factor

Now, we equate the coefficients of this normalized equation with the standard normalized form \(\ddot x + 2\zeta\omega_n\dot x + \omega_n^2 x = 0\):

By comparing the coefficient of the velocity term (\(\dot x\)):

$$ 2\zeta\omega_n = 1.5 $$

By comparing the coefficient of the displacement term (x):

$$ \omega_n^2 = 4.5 $$

First, we determine the natural frequency \(\omega_n\) from the equation for the coefficient of x:

$$ \omega_n = \sqrt{4.5} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} $$

Next, we use the equation for the coefficient of \(\dot x\) to solve for the damping factor \(\zeta\):

$$ \zeta = \frac{1.5}{2\omega_n} $$

Substitute the value of \(\omega_n\) we found:

$$ \zeta = \frac{1.5}{2 \times \frac{3}{\sqrt{2}}} = \frac{1.5}{\frac{6}{\sqrt{2}}} $$

To simplify, multiply the numerator by the reciprocal of the denominator:

$$ \zeta = \frac{1.5 \sqrt{2}}{6} $$

Further simplification yields:

$$ \zeta = \frac{0.5 \sqrt{2}}{2} = 0.25 \sqrt{2} $$

To find the numerical value, we use the approximate value of \(\sqrt{2} \approx 1.41421356\):

$$ \zeta \approx 0.25 \times 1.41421356 \approx 0.35355 $$

Result and Option Matching

The calculated damping factor \(\zeta\) for the given equation of motion is approximately 0.35355.

Comparing this result with the provided options:

• 0.25
• 0.5
• 0.35
• 0.75

The value 0.35355 is closest to 0.35.