Where does the energy vanish in wave interference? [duplicate]

Question

In my physics textbook there is a chapter on waves. Now there is a topic on interference. I don't understand why I get additional energy there. Suppose two waves presented as $Y_1=3\sin(\omega t)$ and $Y_2=3\sin(\omega t)$ now if I add their energy I get total $\text{intensity}=\frac{k}{2}\cdot 3^2+\frac{k}{2}\cdot 3^2$. but again these waves are superimposed if F see this from another viewpoint then I get total $\text{amplitude}=6$. So $\text{energy}=\frac{k}{2}\cdot 6^2$ where $k$ is constant. So where does the energy go? and I tried Googling it, there t found that it's simply rearrangement of energy. But I think that doesn't explain this. And terribly sorry for my bad english

Answer 1

If the waves are from separate sources, in some places they will add together and apparently double the total energy, in other places they will subtract leaving no energy. Elsewhere there will be intermediate situations. Averaged over all space, the energies add correctly, $3^2+3^2$ (ignoring the k/2 factor).
If the two sources are co-located and coherent, the amplitudes will add everywhere in space resulting in double the expected amount of energy, $6^2$. When the two sources are put together, the first source will contribute $3^2$ of that energy, but the second source has to be added on top of the first source and will need more energy, $3*3^2$, to drive the incremental field or voltage. So in this case there will be twice the energy in the field but there has to be twice the energy supplied.

Answer 2

Each individual wave has an energy proportional to its amplitude squared

$$E \propto A^2$$

This means each individual wave with amplitude $3$ has energy proportional to $9$ (ignoring the constant $k$). So the sum total energy of the two separate waves is proportional to $18$ but the waves have not yet added so it makes no sense to square each their amplitudes first and then adding this result to get a total energy.

Now when the waves superimpose, you have a new wave with amplitude $6$ with energy proportional to $6^2$.

Now if we look at the example you have provided, you have two waves called $Y_1$ and $Y_2$ where

$$Y_1 =3\sin(\omega t)$$

$$Y_2 =3\sin(\omega t)$$

You have stated that the sum of their intensities is $3^2 + 3^2 = 18$. This is incorrect. If they were to sum then the resultant intensity would be

$$ \mid Y_1 + Y_2 \mid^2$$

So you would get

$$ \mid Y_1 \mid^2 + 2\mid Y_1 \mid \mid Y_2\mid + \mid Y_2 \mid^2 $$

Once again ignoring the $k^2$ terms, the total intensity would be $3^2 + 2 . 3 . 3 + 3^2 = 36$ consistent with the result above.

Adding the energy of the two initial two waves separately as you have done is not the same as letting the waves superpose and then calculating the energy of the resulting wave. This makes sense if you consider that doubling the amplitude of a wave results in the quadrupling of it’s energy.