The low-frequency speaker of a stereo set has a surface area of 𝐴 equals 0.05 metres squared and produces one watt of acoustical power. What is the intensity at the speaker? If the speaker projects sound uniformly in all directions, at what distance from the speaker is the intensity 0.1 watts per metre squared?
Let’s start by highlighting some of the important information we’ve been given. The speaker has a surface area of 0.05 metres squared called 𝐴. And it produces one watt of acoustical power, which we’ll call 𝑃. We want to know two things: first, the intensity at the speaker, which we’ll call 𝐼 sub 𝑠. And we want to know what distance from the speaker the sound intensity is equal to 0.1 watts per metre squared. We’ll call that distance 𝑑.
Let’s start by drawing a diagram of what’s going on. Here, we show a face on view of the speaker that projects sound out uniformly in all directions. We know the speaker’s area and we know the acoustical power it creates. To solve for the intensity of sound at the speaker, let’s recall a relationship between intensity power and area.
The intensity of sound 𝐼 is equal to the power of the sound produced 𝑃 divided by the area, from which the sound is produced or over which the sound waves fall. In the case of our speaker, the intensity of sound there is equal to the power of sound produced divided by the speaker’s area 𝐴. When we plug in for these two values and calculate this fraction, we find that 𝐼 sub 𝑠 equals 20 watts per metre squared. That’s the sound intensity measured right at the speaker’s surface.
Now, we want to solve for the distance 𝑑 away from the speaker surface that the sound intensity is at a level of 0.1 watts per metre squared. To figure out this distance, we can recall a relationship between intensity 𝐼 and distance from the sound source 𝑟. Sound intensity is proportional to one over the square of the distance.
Now, since we’re told that this speaker emits sound equally in all directions, we can visualize that it emits sound equally in spherical waves travelling forward, backward, up, down, left, right, and in all directions from the speaker. Now, if we were to find the area of one of those spherical waves, that would be the area of a sphere 𝐴 sub 𝑠, which we recall is equal to four times 𝜋 times the sphere’s radius 𝑟 squared.
If we refer back to the equation for sound intensity and apply it to this area of a sphere around the speaker, then intensity 𝐼, which we’re told is 0.1 watts per metre squared at this distance, is equal to the power 𝑃 of the sound produced by the speaker divided by four 𝜋 𝑑 squared.
When we rearrange this equation to solve for 𝑑, we can first multiply both sides by 𝑑 squared, which cancels that term on the right-hand side then divide both sides by 0.1 watts per metre squared, which cancels that turn on the left-hand side. And finally, we take the square root of both sides which cancels the squared term with the square root on the left, leaving us with an equation for 𝑑 that it equals the square root of the power 𝑃 divided by four 𝜋 times 0.1 watts per metre squared.
Plugging in one watt for 𝑃, when we calculate this value, we find that 𝑑 is 0.9 metres. That’s the distance from the speaker surface at which the sound intensity is 0.1 watts per metre squared.