Finding the resonance frequency for forced damped oscillations

Question

I have a problem regarding a forced, damped harmonic oscillator, where I'm trying to find the resonance frequency. I have calculated the frequency for free oscillations as $$\omega_{free}=\sqrt{\frac{\kappa}{I}-\left(\frac{b}{2I}\right)^2},$$ where $b$ is the damping coefficient.
To the best of my knowledge, $\omega_{free}$ should be the same as the resonance frequency, but when I try to calculate the resonance frequency from the amplitude $$A=\frac{\tau_0}{I \sqrt{(\frac{\kappa}{I}-\omega^2)^2 + (\omega \frac{b}{I})^2 }}$$ by finding the maximum value, I get a slightly different equation: $$\omega_{max}=\sqrt{\frac{\kappa}{I}-\left(\frac{\sqrt{2}\cdot b}{2I}\right)^2}.$$ Which one is correct to use as the resonance frequency, and why is $b$ in $\omega_{max}$ scaled by a factor of $\sqrt{2}$ compared to $\omega_{free}$?

Answer 1

Your equations seem to be correct. There are three types of frequencies to consider:

• $\omega_0$ is the frequency of undamped oscillations, i.e. when $b = 0$, aka natural frequency
• $\omega_d$ is the frequency of damped oscillations, i.e. when $0<b<2m\omega_0$
• $\omega_r$ is the frequency at which system gain is maximum, aka resonant frequency

The resonant frequency is not equal to the natural frequency except for undamped oscillators which exist only in theory. Here is a physical (intuitive) explanation:

https://physics.stackexchange.com/a/353061/149541

However, for oscillators with high quality factor the resonant frequency equals natural frequency $\omega_r \approx \omega_0$, as I will show here.

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The differential equation of the forced damped oscillator is:

$$m \ddot{x} + b \dot{x} + k x = u$$

where $m$ is the object mass and $b$ is the dampening coefficient. This system equation is also often written in the following form:

$$\ddot{x} + \gamma \dot{x} + \omega_0^2 x = \frac{1}{m} u$$

where

$$\gamma = \frac{b}{m} \quad \text{and} \quad \omega_0^2 = \frac{k}{m}$$

The quality factor is a dimensionless number that describes how underdamped an oscillator is. The higher the number, the oscillation amplitude decays more slowly:

$$Q = \frac{\omega_0}{\gamma}$$

The transfer function of the system is:

$$G(s) = \frac{1}{m} \frac{1}{s^2 + \gamma s + \omega_0^2} = \frac{1}{m \omega_d} \frac{\omega_d}{(s+\sigma)^2 + \omega_d^2}$$

where

$$\sigma = \frac{\gamma}{2} \quad \text{and} \quad \omega_d = \sqrt{\omega_0^2 - \sigma^2} = \omega_0 \sqrt{1 - \frac{1}{4Q^2}}$$

The system is underdamped when $\omega_0^2 - \sigma^2 > 0$, i.e. when $b < 2 m \omega_0$. When this condition is satisfied the system oscillates with amplitude which decays with time. Also note the effect quality factor has on the system - the higher the $Q$, the oscillations are less damped and the frequency $\omega_d$ is closer to $\omega_0$, where $Q > \frac{1}{2}$.

The response to any input in Laplace domain is $X(s) = G(s) U(s)$. When the input signal is impulse $u(t) = \delta(t) \leftrightarrow U(s) = 1$, then the corresponding response (impulse response) is

$$x(t) = \frac{1}{m\omega_d} e^{-\sigma t} \sin(\omega_d t), \qquad t \geq 0$$

From this it is clear what each parameter does: $\omega_d$ is the frequency of damped oscillations and $\sigma$ is the oscillation amplitude decay rate.

We need to find the transfer function in complex representation:

$$G(j\omega) = \Bigl. G(s) \Bigr|_{s = j\omega} = \frac{1}{m} \frac{1}{(-\omega^2 + \sigma^2 + \omega_d^2) + j(2\sigma\omega)}$$

The system gain is defined as

$$A(\omega) = \left| G(j\omega) \right| = \frac{1}{m} \frac{1}{\sqrt{(\omega^2 - \sigma^2 - \omega_d^2)^2 + (2\sigma\omega)^2}}$$

The maximum gain with respect to frequency can be found from

$$\frac{d}{d\omega} A(w) = -\frac{1}{2m} \frac{2(\omega^2 - \sigma^2 - \omega_d^2)2\omega + 2(2\sigma\omega)2\sigma}{\Bigl(\sqrt{(\omega^2 - \sigma^2 - \omega_d^2)^2 + (2\sigma\omega)^2}\Bigr)^3} = 0$$

The solution is obtained from

$$2(\omega^2 - \sigma^2 - \omega_d^2)2\omega + 2(2\sigma\omega)2\sigma = 0$$

$$\omega^2 = \omega_d^2 - \sigma^2 = \omega_0^2 - \frac{\gamma^2}{2}$$

Therefore, the system gain is at maximum for

$$\omega_r = \sqrt{\omega_d^2 - \sigma^2} = \sqrt{\omega_0^2 - \frac{\gamma^2}{2}} = \omega_0 \sqrt{1 - \frac{1}{2 Q^2}}$$

The resonant frequency equals $\omega_0$ for high-Q oscillators. For example, for $Q = 10$ the resonant frequency is $\omega_r = 0.9975 \cdot \omega_0$.

The system gain at the resonant frequency is

$$\Bigl. A(w) \Bigr|_{\omega=\omega_r} = \frac{1}{m} \frac{1}{2\sigma \omega_d} = \frac{1}{k} \frac{Q}{\sqrt{1 - \frac{1}{4Q^2}}}$$

The system gain is proportional to the Q factor.

Answer 2

Resonant frequency which people define as $\omega_0$ is not the frequency with maximum osccilation, https://youtu.be/Y_DmzZcQR7A Walter lewin explains this at 20:00

$\omega_{max}$ is.