Peak Amplitude for 1 Acoustical Watt Radiated by a Piston into Half-Space as a Function of Radius and Frequency

Abstract

Figure 42 presents solutions to the following equation:

$$ {X_{{peak}}} = (1.18 \times {10^3})({10^{{SPL/20}}})/{f^2}{a^2} $$

(42.1)

where SPL has been fixed at 112 dB, which defines an acoustical power of 1 watt at a distance of 1 meter from a hemispherically radiating source. The peak displacement for any other power P may be determined by multiplying the solution for 1 watt by

$$ {X_{{peak}}} = (1.18 \times {10^3})({10^{{SPL/20}}})/{f^2}{a^2} $$

(42.2)

.