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I think you are using the term "resonance" in a totally unusual and very personal meaning.

What I understand from the expressions you write is your question is that you call "resonance" the value of the frequency for which a particular quantity is maximal for a given input.

If your real question is why is the peak of the acceleration at a higher frequency than the peak of the velocity which itself is at a higher frequency than the peak of the displacement, the answer is given by Roger Wood :

As you point out, amplitude 𝑋(𝜔), peaks at the lowest frequency and then successive derivatives (velocity 𝑗𝜔𝑋(𝜔), acceleration −𝜔2𝑋(𝜔), jerk −𝑗𝜔3𝑋(𝜔), etc.) peak at increasingly higher frequencies because of the successive multiplications by the linear frequency factor 𝑗𝜔.

But the only "resonance" is at $\omega=\omega_0$. Changing the usual definition of a word is generally a bad idea.

EDIT

To turn the perfectly correct answer by Roger Wood into a more "intuitive" one :

The maximum amplitude of the displacement for a given driving force is at some frequency (which happens to be just below $\omega_0$, but this remark is irrelevant to what follows). Since it is a maximum, the slope is zero. If you increase the frequency ever so slightly, the response decreases very, very slowly because there is no downward slope, just a downwards curvature which means you have to build up the down slope by getting away before it starts to go down "seriously". The amplitude of the velocity is the amplitude of the displacement multiplied by $\omega$. If you increase $\omega$ ever so slightly, the product of (displacement times $\omega$) increases by almost the same "ever so slightly" because of the increase $\omega$, since the decrease of the amplitude of the displacement has not yet built up.

So you see the amplitude of the velocity still increases at the "top" of the amplitude of the displacement. Soon enough the decrease of the amplitude of the displacement will build up and dominate the increase due to multiplying by $\omega$, and you will reach the maximum of the velocity for a slightly higher $\omega$ (which happens to be just $\omega_0$ but this is not important.

Repeat exactly the same reasoning, and you'll see that the maximum of the acceleration is still for a slightly higher frequency.

EDIT 2 as an answer to your edit 2

What you call "maximum power absorption" and I would call "maximal dissipation" is obtained at maximal velocity, because it is velocity that causes dissipation, not potential energy.

For exactly the reason I gave in my first edit, maximal velocity is not attained at maximal amplitude of displacement (which is maximal potential energy) but at a slightly (for weak damping) higher frequency.

I think you are using the term "resonance" in a totally unusual and very personal meaning.

What I understand from the expressions you write is your question is that you call "resonance" the value of the frequency for which a particular quantity is maximal for a given input.

If your real question is why is the peak of the acceleration at a higher frequency than the peak of the velocity which itself is at a higher frequency than the peak of the displacement, the answer is given by Roger Wood :

As you point out, amplitude 𝑋(𝜔), peaks at the lowest frequency and then successive derivatives (velocity 𝑗𝜔𝑋(𝜔), acceleration −𝜔2𝑋(𝜔), jerk −𝑗𝜔3𝑋(𝜔), etc.) peak at increasingly higher frequencies because of the successive multiplications by the linear frequency factor 𝑗𝜔.

But the only "resonance" is at $\omega=\omega_0$. Changing the usual definition of a word is generally a bad idea.

EDIT

To turn the perfectly correct answer by Roger Wood into a more "intuitive" one :

The maximum amplitude of the displacement for a given driving force is at some frequency (which happens to be just below $\omega_0$, but this remark is irrelevant to what follows). Since it is a maximum, the slope is zero. If you increase the frequency ever so slightly, the response decreases very, very slowly because there is no downward slope, just a downwards curvature which means you have to build up the down slope by getting away before it starts to go down "seriously". The amplitude of the velocity is the amplitude of the displacement multiplied by $\omega$. If you increase $\omega$ ever so slightly, the product of (displacement times $\omega$) increases by almost the same "ever so slightly" because of the increase $\omega$, since the decrease of the amplitude of the displacement has not yet built up.

So you see the amplitude of the velocity still increases at the "top" of the amplitude of the displacement. Soon enough the decrease of the amplitude of the displacement will build up and dominate the increase due to multiplying by $\omega$, and you will reach the maximum of the velocity for a slightly higher $\omega$ (which happens to be just $\omega_0$ but this is not important.

Repeat exactly the same reasoning, and you'll see that the maximum of the acceleration is still for a slightly higher frequency.

EDIT 2 as an answer to your edit 2

What you call "maximum power absorption" and I would call "maximal dissipation" is obtained at maximal velocity, because it is velocity that causes dissipation, not potential energy.

For exactly the reason I gave in my first edit, maximal velocity is not attained at maximal amplitude of displacement (which is maximal potential energy) but at a slightly (for weak damping) higher frequency.

EDIT 3

In reaction to Farcher's comment : I had assumed you considered an electrical circuit. This is why I said the usual definition of resonance is $\omega=\omega_0$, maximal velocity.

For Mechanics, indeed, the usual the usual definition of resonance is maximal displacement. This is indeed an inconsistency, due to an old historical situation.

But my point is, in a given setting, whether Mechanics or electricity, there is one usual definition of resonance. So keep to the usual definition in that whatever setting your are dealing with, even if there is another usual definition for a different setting.

I think you are using the term "resonance" in a totally unusual and very personal meaning.

What I understand from the expressions you write is your question is that you call "resonance" the value of the frequency for which a particular quantity is maximal for a given input.

If your real question is why is the peak of the acceleration at a higher frequency than the peak of the velocity which itself is at a higher frequency than the peak of the displacement, the answer is given by Roger Wood :

As you point out, amplitude 𝑋(𝜔), peaks at the lowest frequency and then successive derivatives (velocity 𝑗𝜔𝑋(𝜔), acceleration −𝜔2𝑋(𝜔), jerk −𝑗𝜔3𝑋(𝜔), etc.) peak at increasingly higher frequencies because of the successive multiplications by the linear frequency factor 𝑗𝜔.

But the only "resonance" is at $\omega=\omega_0$. Changing the usual definition of a word is generally a bad idea.

EDIT

To turn the perfectly correct answer by Roger Wood into a more "intuitive" one :

The maximum amplitude of the displacement for a given driving force is at some frequency (which happens to be just below $\omega_0$, but this remark is irrelevant to what follows). Since it is a maximum, the slope is zero. If you increase the frequency ever so slightly, the response decreases very, very slowly because there is no downward slope, just a downwards curvature which means you have to build up the down slope by getting away before it starts to go down "seriously". The amplitude of the velocity is the amplitude of the displacement multiplied by $\omega$. If you increase $\omega$ ever so slightly, the product of (displacement times $\omega$) increases by almost the same "ever so slightly" because of the increase $\omega$, since the decrease of the amplitude of the displacement has not yet built up.

So you see the amplitude of the velocity still increases at the "top" of the amplitude of the displacement. Soon enough the decrease of the amplitude of the displacement will build up and dominate the increase due to multiplying by $\omega$, and you will reach the maximum of the velocity for a slightly higher $\omega$ (which happens to be just $\omega_0$ but this is not important.

Repeat exactly the same reasoning, and you'll see that the maximum of the acceleration is still for a slightly higher frequency.

I think you are using the term "resonance" in a totally unusual and very personal meaning.

What I understand from the expressions you write is your question is that you call "resonance" the value of the frequency for which a particular quantity is maximal for a given input.

If your real question is why is the peak of the acceleration at a higher frequency than the peak of the velocity which itself is at a higher frequency than the peak of the displacement, the answer is given by Roger Wood :

As you point out, amplitude 𝑋(𝜔), peaks at the lowest frequency and then successive derivatives (velocity 𝑗𝜔𝑋(𝜔), acceleration −𝜔2𝑋(𝜔), jerk −𝑗𝜔3𝑋(𝜔), etc.) peak at increasingly higher frequencies because of the successive multiplications by the linear frequency factor 𝑗𝜔.

But the only "resonance" is at $\omega=\omega_0$. Changing the usual definition of a word is generally a bad idea.

EDIT

To turn the perfectly correct answer by Roger Wood into a more "intuitive" one :

The maximum amplitude of the displacement for a given driving force is at some frequency (which happens to be just below $\omega_0$, but this remark is irrelevant to what follows). Since it is a maximum, the slope is zero. If you increase the frequency ever so slightly, the response decreases very, very slowly because there is no downward slope, just a downwards curvature which means you have to build up the down slope by getting away before it starts to go down "seriously". The amplitude of the velocity is the amplitude of the displacement multiplied by $\omega$. If you increase $\omega$ ever so slightly, the product of (displacement times $\omega$) increases by almost the same "ever so slightly" because of the increase $\omega$, since the decrease of the amplitude of the displacement has not yet built up.

So you see the amplitude of the velocity still increases at the "top" of the amplitude of the displacement. Soon enough the decrease of the amplitude of the displacement will build up and dominate the increase due to multiplying by $\omega$, and you will reach the maximum of the velocity for a slightly higher $\omega$ (which happens to be just $\omega_0$ but this is not important.

Repeat exactly the same reasoning, and you'll see that the maximum of the acceleration is still for a slightly higher frequency.

EDIT 2 as an answer to your edit 2

What you call "maximum power absorption" and I would call "maximal dissipation" is obtained at maximal velocity, because it is velocity that causes dissipation, not potential energy.

For exactly the reason I gave in my first edit, maximal velocity is not attained at maximal amplitude of displacement (which is maximal potential energy) but at a slightly (for weak damping) higher frequency.

I think you are using the term "resonance" in a totally unusual and very personal meaning.

What I understand from the expressions you write is your question is that you call "resonance" the value of the frequency for which a particular quantity is maximal for a given input.

If your real question is why is the peak of the acceleration at a higher frequency than the peak of the velocity which itself is at a higher frequency than the peak of the displacement, the answer is given by Roger Wood :

As you point out, amplitude 𝑋(𝜔), peaks at the lowest frequency and then successive derivatives (velocity 𝑗𝜔𝑋(𝜔), acceleration −𝜔2𝑋(𝜔), jerk −𝑗𝜔3𝑋(𝜔), etc.) peak at increasingly higher frequencies because of the successive multiplications by the linear frequency factor 𝑗𝜔.

But the only "resonance" is at $\omega=\omega_0$. Changing the usual definition of a word is generally a bad idea.

I think you are using the term "resonance" in a totally unusual and very personal meaning.

What I understand from the expressions you write is your question is that you call "resonance" the value of the frequency for which a particular quantity is maximal for a given input.

If your real question is why is the peak of the acceleration at a higher frequency than the peak of the velocity which itself is at a higher frequency than the peak of the displacement, the answer is given by Roger Wood :

As you point out, amplitude 𝑋(𝜔), peaks at the lowest frequency and then successive derivatives (velocity 𝑗𝜔𝑋(𝜔), acceleration −𝜔2𝑋(𝜔), jerk −𝑗𝜔3𝑋(𝜔), etc.) peak at increasingly higher frequencies because of the successive multiplications by the linear frequency factor 𝑗𝜔.

But the only "resonance" is at $\omega=\omega_0$. Changing the usual definition of a word is generally a bad idea.

EDIT

To turn the perfectly correct answer by Roger Wood into a more "intuitive" one :

The maximum amplitude of the displacement for a given driving force is at some frequency (which happens to be just below $\omega_0$, but this remark is irrelevant to what follows). Since it is a maximum, the slope is zero. If you increase the frequency ever so slightly, the response decreases very, very slowly because there is no downward slope, just a downwards curvature which means you have to build up the down slope by getting away before it starts to go down "seriously". The amplitude of the velocity is the amplitude of the displacement multiplied by $\omega$. If you increase $\omega$ ever so slightly, the product of (displacement times $\omega$) increases by almost the same "ever so slightly" because of the increase $\omega$, since the decrease of the amplitude of the displacement has not yet built up.

So you see the amplitude of the velocity still increases at the "top" of the amplitude of the displacement. Soon enough the decrease of the amplitude of the displacement will build up and dominate the increase due to multiplying by $\omega$, and you will reach the maximum of the velocity for a slightly higher $\omega$ (which happens to be just $\omega_0$ but this is not important.

Repeat exactly the same reasoning, and you'll see that the maximum of the acceleration is still for a slightly higher frequency.