声强
每单位面积传递给介质的功率，其中该面积垂直于波的传播方向。要使声强具有意义，必须指明测量位置相对于声源的位置。介质通常是流体。（Kinsler，第 110 页）
如果指定位置位于声源表面，则声源与介质之间界面处的平均声强为 —
\begin{align} \label{eq:11301a} \textsf{Acoustic intensity (average)} = \frac{\textsf{Power delivered to the medium}}{\textsf{Area from which the power is delivered}} \end{align}
对于给定的谐振器材料和负载，在磨损机理相同的前提下，较高的声源声强会导致声源谐振器更大的磨损。
以下示例展示了逐渐增大的声强水平。
示例 1
一台 20 kHz 超声清洗器的有效面积为 100 mm x 100 mm。在水负载下，传递的功率为 75 瓦。平均声强为 —
\begin{align} \label{eq:11302a} \textsf{Acoustic intensity} &= \frac{\textsf{75 watts}}{\textsf{10000 mm}^2} \\[0.7em]%eqn_interline_spacing &= \textsf{0.0075 watts/mm}^2 \nonumber \end{align}
示例 2
一根 20 kHz 圆柱形变幅杆（端面直径 10 mm），振幅为 100 微米。当端面浸入水中时，传递的功率为 200 瓦。平均声强为 —
\begin{align} \label{eq:11303a} \textsf{Acoustic intensity} &= \frac{\textsf{200 watts}}{\textsf{79 mm}^2} \\[0.7em]%eqn_interline_spacing &= \textsf{2.5 watts/mm}^2 \nonumber \end{align}
示例 3
一根 20 kHz 金属焊接变幅杆（接触面积 8 mm x 8 mm），振幅为 100 微米。在焊接过程中，传递的功率为 900 瓦。声强为 —
\begin{align} \label{eq:11304a} \textsf{Acoustic intensity} &= \frac{\textsf{900 watts}}{\textsf{64 mm}^2} \\[0.7em]%eqn_interline_spacing &= \textsf{14 watts/mm}^2 \nonumber \end{align}
Acoustic intensity
The power per unit area that is delivered to a medium, where the area is perpendicular to the direction of wave travel. To have meaning, the location where the intensity is measured must be specified with respect to the acoustic source. The medium is often a fluid. (Kinsler, p. 110)
If the specified location is at the surface of the acoustic source, then the average acoustic intensity at the interface between the source and the medium is —
\begin{align} \label{eq:11301a} \textsf{Acoustic intensity (average)} = \frac{\textsf{Power delivered to the medium}}{\textsf{Area from which the power is delivered}} \end{align}
For a given resonator material and load, a higher source acoustic intensity will result in greater wear of the source resonator, assuming that the wear mechanism is the same.
The following examples illustrate increasing levels of acoustic intensity.
Example 1
A 20 kHz ultrasonic cleaner has an active area of 100 mm x 100 mm. With a water load, the delivered power is 75 watts. The average acoustic intensity is —
\begin{align} \label{eq:11302a} \textsf{Acoustic intensity} &= \frac{\textsf{75 watts}}{\textsf{10000 mm}^2} \\[0.7em]%eqn_interline_spacing &= \textsf{0.0075 watts/mm}^2 \nonumber \end{align}
Example 2
A 20 kHz cylindrical horn (10 mm face diameter) has an amplitude of 100 microns. When the face is immersed in water, 200 watts of power are delivered. The acoustic average intensity is —
\begin{align} \label{eq:11303a} \textsf{Acoustic intensity} &= \frac{\textsf{200 watts}}{\textsf{79 mm}^2} \\[0.7em]%eqn_interline_spacing &= \textsf{2.5 watts/mm}^2 \nonumber \end{align}
Example 3
A 20 kHz metal welding horn (8 mm x 8 mm contact area) has an amplitude of 100 microns. During welding, 900 watts of power are delivered. The acoustic intensity is —
\begin{align} \label{eq:11304a} \textsf{Acoustic intensity} &= \frac{\textsf{900 watts}}{\textsf{64 mm}^2} \\[0.7em]%eqn_interline_spacing &= \textsf{14 watts/mm}^2 \nonumber \end{align}