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 $\omega_{free}$?
Finding the resonance frequency for forced damped oscillations
dissipationfrequencyharmonic-oscillatoroscillatorsresonance
Best Answer
Your equations seem to be correct. There are three types of frequencies to consider:
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.
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.