Question

A sound wave with intensity \(25 \mathrm{mW} / \mathrm{m}^{2}\) interferes destructively with a sound wave that has an intensity of $28 \mathrm{mW} / \mathrm{m}^{2} .$ What is the intensity of the superposition of the two?

Short answer

Answer: The intensity of the superposition of the two sound waves is 3 mW/m².

Step-by-step solution

Identify the given intensities

We are given the intensities of two sound waves: Intensity 1: \(I_1 = 25 \mathrm{mW} / \mathrm{m}^{2}\) Intensity 2: \(I_2 = 28 \mathrm{mW} / \mathrm{m}^{2}\)

Use the principle of superposition for destructive interference

The principle of superposition states that when two waves interfere destructively, we should subtract the individual intensities to obtain the final intensity. So, we will calculate the resulting intensity using this formula: \(I_r = |I_1 - I_2|\)

Calculate the final intensity

Now, we will plug the values of the intensities into the formula and calculate the final intensity of the superimposed sound waves: \(I_r = |25\mathrm{mW}/\mathrm{m}^2 - 28 \mathrm{mW}/\mathrm{m}^2|\) \(I_r = |(-3) \mathrm{mW}/\mathrm{m}^2|\) \(I_r = 3 \mathrm{mW}/\mathrm{m}^2\) The intensity of the superposition of the two sound waves is \(3\mathrm{mW}/\mathrm{m}^2\).