In order for the intensity of a light source to stay the same, while each lower frequency photon carries less energy, there must be a greater number (per time, per area) of the lower frequency photons in the beam than the original number of higher frequency photons.
As for the second part of your question, I admit that it can be confusing that the power transmitted by E&M waves depends on the amplitude of the wave, while the power transmitted by a mode of a vibrating string depends on both the amplitude and the frequency of the wave. Ultimately this comes down to fundamental differences in the physics of each wave phenomenon.
The energy in a vibrating string is reducible to the kinetic energy of the moving string elements and the potential energy from the tension felt by each element due to the position of its neighbors. So, at fixed amplitude, you can see that you get even more energy if you jiggle the rope faster.
The energy in an E&M wave is a different effect entirely: it comes from the average size of the (squared) electric field in the wave that can do work to move charged particles. At a fixed amplitude, if you increase the frequency you won't increase the average size of the field.
The energy of an oscillator, as Planck defined, is quantized. It may be related to a harmonic oscillator, but it won't be simple to explain whose is the potential energy and the kinetic energy. We don't have here some sort of mass in some potential field, and the mass moves and has potential and kinetic energy. We have here the field itself and it is a thing that carries an energy of its own. So, the energy of the e.m. field is defined otherwise, not with potential and kinetic energy,
$ (\text I) \ H = \frac {1}{2} \int (\epsilon _0 |\vec E|^2 + \frac {1}{\mu _0} |\vec H|^2) \text d \vec r$.
where $H$ is the energy (the Hamiltonian). The vectors $\vec E$ and $\vec H$ also have an amplitude that oscillates in time.
Now, in the quantum theory, the fields $\vec E$ and $\vec H$, are replaced by the operators $\hat {\vec E}$ and $\hat {\vec H}$, (see D. F. Walls and G. J. Milburn, "Quantum Optics"). So, in the Hamiltonian $(\text I)$ the squared amplitudes $|\vec E|^2$ and $|\vec H|^2$ become operators. Next, to get the energy of the photon(s) in some state $|\psi\rangle$ of the e.m. field, we calculate the average of the Hamiltonian operator obtained, in that state. $ \langle \psi | \hat H|\psi \rangle$, and you will see that what we obtain is that the energy is given by Plank's formula.
I will show you a few steps of such a calculus.
a) In the classical electromagnetism $\vec E$ and $\vec B$ are obtained from the vector potential
$ (\text {II}) \hat E = - \frac {∂ \hat A}{∂t}, \ \ \ \hat B = \nabla \times \vec A .$
The same we do here, we define an operator vector potential
$ (\text {III}) \hat {\vec A(r,t)} = i \sum _k ( \frac {\hbar }{2 \omega _k \epsilon _0})^{1/2}[\hat a_k u_k(\vec r) e^{-i\omega _k t} + \hat a^{\dagger}_k u^*_k(\vec r) e^{i\omega _k t}]$,
where the index $k$ is for the $k$-th frequency (more exactly angular velocity) in the field, i.e. we number $\omega_1, \omega_2, ... \omega_k,...$; $C_k$ is a constant depending on $\omega_k$, and $a_k, a^{\dagger}_k$ are the annihilation and creation operators.
b) From $\hat {\vec A(r,t)} $ we calculate $\hat {\vec E}$ and $\hat {\vec B}$ applying the formulas in $ (\text {II})$. Then we take their absolute squares, i.e $\hat {\vec E}^{\dagger} \cdot \hat {\vec E}$ and $\hat {\vec B}^{\dagger} \cdot \hat {\vec B}$.
c) We introduce these vectors in the Hamiltonian, i.e.
$ (\text IV) \ \hat H = \frac {1}{2} \int (\epsilon _0 \hat {\vec E}^{\dagger} \hat {\vec E} + \frac {1}{\mu _0} \hat {\vec B}^{\dagger} \hat {\vec B} ) \text d \vec r$.
Performing the integral and applying different rules of the annihilation and creation operators, we get
$ (\text {V}) \ \hat H = \sum _k \hbar \omega _k (\hat a^{\dagger} \hat a + \frac{1}{2})$.
d) Now, as I said, we calculate the average for the state given for the e.m. field. Let's take a simple state, $|n\rangle$, which means that we have $n$ photons of the same frequency $\nu = \omega/2\pi$. We get the energy
$ (\text {VI}) \ \mathscr E = \langle \hat {H} \rangle = (n + \frac {1}{2})\hbar \omega$
in agreement with Planck's formula up to the term $\hbar \omega /2$ that I won't discuss here.
Best Answer
Let me say what others are trying to say, hopefully in a clearer fashion:
Just because you can relate two variables in an equation does not mean that they are dependant. In this case, you have to constrain intensity $I$ in order to get the relationship. At that point, it is not a general relationship, but only true when $I$ is constrained.
An example that might be easier to see intuitively would be: $$KE=\frac{1}{2}mv^2$$ If you constrain kinetic energy you can get a relationship between mass and velocity. For example: $$m=2\frac{KE}{v^2}$$ But intuitively, you know that mass and velocity are independent of one another. Why would changing the mass of an object inherently change the velocity? But, if the kinetic energy is held constant, then it would force a relationship between them. A relationship that is not generally meaningful.
So, to bring this back to your case, $x$ sound amplitude and $w$ angular frequency are independent of each other, but you can force a relationship between them by constraining $I$, but it is not a meaningful or general relationship.
I found a good answer elsewhere that explains it much better than I did here. I would recommend checking it out.