Question

Power of a given frequency is proportional to the square of the amplitude

Answer 1

It's not true in general that the energy of a wave is always proportional to the square of its amplitude, but there are good reasons to expect this to be true in most cases, in the limit of small amplitudes. This follows simply from expanding the energy in a Taylor series, E=a0+a1A+a2A2+…E=a0+a1A+a2A2+… We can take the a0a0 term to be zero, since it would just represent some potential energy already present in the medium when there was no wave excitation. The a1a1 term has to vanish, because otherwise it would dominate the sum for sufficiently small values of AA, and you could then have waves with negative energy for an appropriately chosen sign of AA. That means that the first nonvanishing term should be A2A2. Since we don't expect the energy of the wave to depend on phase, we expect that only the even terms should occur, E=a2A2+a4A4+…E=a2A2+a4A4+… So it's only in the limit of small amplitudes that we expect E∝A2E∝A2.

The other issue to consider is that we had to assume that EE was a sufficiently smooth function of AA to allow it to be calculated using a Taylor series. This doesn't have to be true in general. As an easy example involving an oscillating particle, rather than a wave, consider a pointlike particle in a gravitational field, bouncing up and down elastically on an inflexible floor. If we define the amplitude as the height of the bounce, then we have E∝|A|E∝|A|. But a realistic ball deforms, so the small-amplitude limit consists of the ball vibrating while remaining in contact with the floor, and we regain E∝A2E∝A2.

You could also make up examples where a2a2 vanishes and the first nonvanishing coefficient is a4a4.

hope it will help you