# Intensity and amplitude

1. Aug 14, 2013

### Gauss M.D.

This is something I don't relly get. I keep reading that the intensity of a wave is proportional to the square of its amplitude.

So let's suppose we have a random wave source, with amplitude A0. If we replace that source with five identical copies of it, their collective amplitude is 5A, yes? But have I really raised the energy level at the source by a factor of 25? What am I missing here?

(X-posting since I think the subforum I posted in was not appropriate)

2. Aug 14, 2013

### barryj

Lets look at an electrical example. Power = (voltage)^2/R. In general power is equivalent to intensity while voltage is equivalent of amplitude. If you double the voltage, the power goes up by four. If voltage goes up by 5, power goes up by 25.

3. Aug 14, 2013

### tiny-tim

i'll use the electrical example slightly differently …

the energy (or power) is proportional to VI (voltage times current) …

how are you going to double the current without doubling the voltage?

4. Aug 15, 2013

### rude man

If they are really copies of each other, the amplitudes will add and the intensity is indeed 25x.

But they must be in phase with each other.

5. Aug 15, 2013

### haruspex

Consider two sources like sin(t). (Sound waves, say, except that I'll ignore attenuation with distance.) Suppose they are not co-located, so in places they will interfere constructively and in other places destructively. In general, they may be received as sin(t+α) and sin(t-α). That adds to 2 sin(t)cos(α). Squaring and integrating 0 to 2π gives 4π cos2(α). If we suppose all phase differences (α values) occur equally across a region, we can integrate wrt α and obtain an average value of 2π. A single sin(t) source gives π, so the power from two sources is double, as expected.
If the sources are exactly co-located then indeed you will get four times the power everywhere, but in this case the source is working four times as hard.