Question

A second order system has $\omega_n = 4\text{ rad/sec}$ and $\zeta = 0.5$. What is the damped natural frequency $\omega_d$ ?

• A. $4$
• B. $2$
• C. $1.73$
• D. $3.46$

Hint:

For an underdamped second-order system, remember the right-angled triangle relationship: the hypotenuse is $\omega_n$, the horizontal side is the attenuation factor $\sigma = \zeta\omega_n$, and the vertical side is $\omega_d$. Thus, $\omega_d = \omega_n\sqrt{1-\zeta^2}$.

Answer 1

The Correct Option is D

Concept: In second-order system dynamics, when the damping ratio satisfies $0 < \zeta < 1$, the system is underdamped. The transient oscillations occur at a frequency lower than the undamped natural frequency ($\omega_n$), known as the damped natural frequency ($\omega_d$). The relationship is defined by: \[ \omega_d = \omega_n \sqrt{1 - \zeta^2} \]

Step 1: Substitute the given parameters into the formula.
The problem provides the following parameters: - Undamped natural frequency: $\omega_n = 4\text{ rad/sec}$ - Damping ratio: $\zeta = 0.5$ Substitute these values into the expression for $\omega_d$: \[ \omega_d = 4 \times \sqrt{1 - (0.5)^2} \]

Step 2: Evaluate the mathematical expression.
First, calculate the square of the damping ratio: \[ (0.5)^2 = 0.25 \] Subtract this from 1 inside the radical: \[ 1 - 0.25 = 0.75 = \frac{3}{4} \] Now evaluate the square root: \[ \sqrt{0.75} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] Substitute this back into the equation for $\omega_d$: \[ \omega_d = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \] Using the decimal approximation for $\sqrt{3} \approx 1.73205$: \[ \omega_d = 2 \times 1.73205 = 3.4641\text{ rad/sec} \] Rounding to two decimal places yields $3.46$.