Two loudspeakers on elevated platforms are at opposite ends

Two loudspeakers on elevated platforms are at opposite ends of a field. Each broadcasts equally in all directions. The sound intensity level at a point halfway between the loudspeakers is 71.6 dB .

Q:What is the sound intensity level at a point one-quarter of the way from one speaker to the other along the line joining them?

Solution

We know that the net intensity level due to multiple source is given as: 71.6 = 10 log ( Inet / I₀);

71.6 = 10 log(2I/Io)

Now, what we have is that the Inet = 2I where I is the intensity due to one of the speakers.

[Since the two loudspeakers are the same, we take Intensity at the centre of the field due be same]

Further, we know that Intensity due to a sound source is inversely proportional to the distance from the source.

That is the net intensity at a point one quarter away from one of the sources = 4I + 4I/9 = 40 I/9

[Since the distance from one of the speakers will be halved, the intensity becomes 4 times, while for other the distance will become 3/2 times, hence the intensity will become 4/9 times]

Hence, the new sound intensity level = 10 log(40 I / 9 Io) = 10log [(20/9)*(2I/Io)] = 10log(20/9) + 71.6

Therefore, the net sound intensity level = 75.0679 dB