# Mathematical demonstration

It's straightforward to see why this happens if you use a bit of linear response theory.
Consider a generic damped harmonic oscillator.
There are three forces, the restoring force $F_\text{restoring} = - k x(t)$, the friction force $F_\text{friction} = - \mu \dot{x}(t)$, and the driving force $F_\text{drive}(t)$.
Newton's law says $F(t) = m \ddot{x}(t)$ which gives
$$-k x(t) - \mu \dot{x}(t) + F_\text{drive}(t) = m \ddot{x}(t) \, .$$
Dividing through by $m$ and defining $\phi(t) \equiv x(t)/m$, $\omega_0^2 \equiv k/m$, $2 \beta \equiv \mu/m$, and $J(t) \equiv F_\text{drive}(t)/m$, we get
$$ \ddot{\phi}(t) + 2\beta \dot{\phi}(t) + \omega_0^2 \phi(t) = J(t) \, .$$
This is a nice general form of the damped driven harmonic oscillator.

Writing $\phi(t)$ as a Fourier transform
$$\phi(t) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \, \tilde{\phi}(\omega) e^{i \omega t}$$
and plugging into the equation of motion, we find
$$\left( - \omega^2 + i 2 \beta \omega + \omega_0^2 \right) = \tilde{J}(\omega)$$
which can be rewritten as
$$\tilde{\phi}(\omega) = - \frac{\tilde{J}(\omega)}{\left( \omega^2 - i 2 \beta \omega - \omega_0^2 \right)} \, .$$

Let's take the case where the drive is a cosine, i.e. $J(t) = A \cos(\Omega t)$.
In this case $\tilde{J}(\omega) = (1/2)\left(\delta(\omega - \Omega) + \delta(\omega + \Omega) \right)$ so if you work it all out you find
$$\phi(t) = \Re \left[ - \frac{A e^{i \Omega t}}{\Omega^2 - i 2 \beta \Omega - \omega_0^2} \right] \, .$$
It's easy to check that $\phi(t)$ has the largest amplitude when $\Omega = \omega_0 \sqrt{1 - 2 (\beta / \omega_0)^2}$, which decreases as $\beta$ increases.
Remember that $\beta$ is just proportional to the coefficient of friction $\mu$ so we've shown that more friction makes the peak move to lower frequency.

# Resonance

We've shown that the amplitude of the oscillator depends on the damping coefficient.
However, this does *not* mean that the resonance moves to lower frequency.
Resonance is a condition *defined* by unidirectional flow of energy from the drive to the system.
It turns out (easy to show with the math we already did) that this happens when $\Omega = \omega_0$, i.e. the drive is at the same frequency as the *undamped* oscillation frequency.
There's already a nice post on this issue which I recommend reading.

# Original questions

Why does this happen?

Well, we showed why mathematically.
Intuitively it's because the friction takes away kinetic energy so the oscillator doesn't make it as far from equilibrium on each cycle.

And does this imply that at higher damping levels one cannot achieve a higher amplitude by setting the period of the forced oscillation to be equal to the natural frequency?

Assuming constant amplitude of the drive, yes.

This seems strange because, the highest amplitude is achieved when the natural frequency is equal to the driving frequency. But, I guess that the rules are somehow different for the damping case.

Indeed, damping changes things a bit.

# Other reading

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.

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.

## Best Answer

Resonanceis when you push a kid on a swing.If you do it right, you make him swing higher (You push him in the same direction as he is moving). If you do it wrong, you stop his swinging (e.g. pushing, when he is on the way backwards).

When you "add to" or "build up" the amplitude of the oscillations,

this is resonance.You need to push with the same frequency as he is swinging. (If he is swinging back-and-forth in 3 seconds, then you must push forward exactly once every 3 seconds). Also you must be in phase. (Even when your push-frequency is equal to his swing-frequency, you should not push when he is moving backwards.)

A damped oscillation only changes amplitude with time. Not frequency. Resonance is not different in this case.

If you want a steady oscillation with constant amplitude, then of course you need to force an oscillation on the system and reach resonance to cancel out the damping.