Example 4: Determining the Resonance Frequency for an Absorber of Specified Dimensions.
Suppose that we have a sheet of 16 gauge sheet metal, perforated with Y8-in holes, staggered at 3/8-in on-center (about 8 holes/sq in), which is used as a facing for a glass wool blanket 3/4-in thick, against a solid wall. Determine the resonance frequency.
For this example:
b = 0.375";
d = 0.125";
t = 0.0625";
h = 0.75";
e = 0.0625 + 0.8 x 0.125
= 0.1625;
p = 0.9 (0.125"/0.375")2 x 100
= 10%
We begin by locating the points on the nomogram corresponding to e = 0.16" and p = 10% and connecting these points with a straight line. Mark the point where this line crosses the unnumbered "m"-scale. Now connect that point with the point on the "h"-scale corresponding to the absorber depth, h = 0.75". Read the resonance frequency where this line crosses the fR-scale: 2000 Hz.
This would be a suitable structure for the jet engine duct lining mentioned above .
