4 Engineering Example 2

4.1 Exponential decay of sound intensity


The rate at which a quantity decays is important in many branches of engineering and science. A particular example of this is exponential decay. Ideally the sound level in a room where there are substantial contributions from reflections at the walls, floor and ceiling will decay exponentially once the source of sound is stopped. The decay in the sound intensity is due to absorbtion of sound at the room surfaces and air absorption although the latter is significant only when the room is very large. The contributions from reflection are known as reverberation . A measurement of reverberation in a room of known volume and surface area can be used to indicate the amount of absorption.

Problem in words

As part of an emergency test of the acoustics of a concert hall during an orchestral rehearsal, consultants asked the principal trombone to play a single note at maximum volume. Once the sound had reached its maximum intensity the player stopped and the sound intensity was measured for the next 0.2 seconds at regular intervals of 0.02 seconds. The initial maximum intensity at time 0 was 1. The readings were as follows:

time 0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
intensity 1 0.63 0.35 0.22 0.13 0.08 0.05 0.03 0.02 0.01 0.005

Draw a graph of intensity against time and, assuming that the relationship is exponential, find a function which expresses the relationship between intensity and time.

Mathematical statement of problem

If the relationship is exponential then it will be a function of the form

I = I 0 1 0 k t

and a log-linear graph of the values should lie on a straight line. Therefore we can plot the values and find the gradient and the intercept of the resulting straight-line graph in order to find the values for I 0 and k .

k is the gradient of the log-linear graph i.e.

k = change in log 10 (intensity) change in time

and I 0 is found from where the graph crosses the vertical axis log 10 ( I 0 ) = c

Mathematical analysis

Figure 9(a) shows the graph of intensity against time.

We calculate the log 10 (intensity) to create the table below:

time 0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
log 10 (intensity) 0 -0.22 -0.46 -0.66 -0.89 -1.1 -1.3 -1.5 -1.7 -2.0 -2.2

Figure 9(b) shows the graph of log (intensity) against time.

Figure 9 :

{(a) Graph of sound intensity against time (b) Graph of $log_{10}$ (intensity) against time and a line fitted by eye to the data. The line goes through the points $(0,0)$ and $(0.2, -2.2)$.}

We can see that the second graph is approximately a straight line and therefore we can assume that the relationship between the intensity and time is exponential and can be expressed as

I = I 0 1 0 k t .

The log 10 of this gives

log 10 ( I ) = log 10 ( I 0 ) + k t .

From the graph (b) we can measure the gradient, k using

k = change in log 10 (intensity) change in time

giving k = 2.2 0 0.2 0 = 11

The point at which it crosses the vertical axis gives

log 10 ( I 0 ) = 0 I 0 = 1 0 0 = 1

Therefore the expression I = I 0 1 0 k t becomes

I = 1 0 11 t


The data recorded for the sound intensity fit exponential decaying with time. We have used a log-linear plot to obtain the approximate function:

I = 1 0 11 t