Sound travel problem

Question

Here's an example of my question to make my explanation a bit easier.

Say a decent loudspeaker plays a tune at loud volume 100m away from me and another speaker plays the same tune at lot lower volume 1m away from me. Say the tunes will be just as loud in my ears, and if I'm understanding this correctly we could measure the decibel and record the sounds and they would be equal as well (if we ignore the fact that certain wavelengths travel better through air than others). But a person standing a further 100m away would only hear the loudest speaker (now 200m away), and not the low volume one (101m away). The dB and the sounds wouldn't be identical anymore.To me this doesn't make sense as sound is just air, and I would believe equally loud sounds at one place would fade equally with the same distance added.

So my question, why am I wrong (why isn't it like this in reality)?

Please forgive me for my explanatory problems as English is not my native language.

Answer 1

We can understand it in the following way, where the key is that at different positions the energies of the sound are different.

Suppose a speaker generates sound wave with energy $A$ at its own position, by energy conservation, at distance $R$, the energy one receives per unit area is

\begin{equation} A/(4\pi R^2) \end{equation}

Here I assume the space is isotropic in all directions.

Now with the two speakers, suppose the louder speaker generates sound wave with energy $A_1$ while the quieter speaker generates sound wave with energy $A_2$. Here both $A_1$ and $A_2$ are the energies generated at the positions of the two speakers, respectively. Again, by energy conservation and assuming the space is isotropic, we have

\begin{split} A_1/(4\pi R_1^2)=A_2/(4\pi R_2^2) \end{split} because these two speakers sound the same loud. When $R_1=100R_2$, we have

\begin{split} A_1/A_2=10000 \end{split}

Now for an observer who stands at a position which is $200{\rm m}$ from the loud speaker and $101{\rm m}\approx100{\rm m}$, the energy produced by the loud speaker is \begin{equation} A_1/(4\pi\times (200{\rm m})^2) \end{equation} and that produced by the quieter speaker is approximately \begin{equation} A_2/(4\pi\times (100{\rm m})^2) \end{equation}

Because $A_1/A_2=10000$, this time

\begin{equation} \frac{A_1/(4\pi\times (200{\rm m})^2)}{A_2/(4\pi\times (100{\rm m})^2)}=2500 \end{equation} which means the energy from the louder speaker at this position is much larger than that from the quieter speaker, that is why that person can only hear the louder speaker.

Answer 2

Simply - inverse square law. Forgetting for a moment that you are standing on a surface (reflections complicate things), all the energy from the loudspeaker is traveling outward in a sphere.

At the point where you first measured, one sphere has a radius of 1 m, the other a radius of 100 m. At this point amplitudes are the same - implying that energy per unit area is the same.

When you go another 100 m further, the first sphere has grown to 101 m radius: the energy per unit area has dropped by $(1/101)^2$. By contrast, the second sphere has doubled in size, so the area has quadrupled: its energy per unit area is now $1/4$ of what it was before. This is why the second loudspeaker's sound sounds much louder when you move further away.