Everything Waves: Sound


Power in physics

Centripetal acceleration

Series circuits


Left hand rules

Properties of magnets and motors


Newton's Law of Gravitation

Strong Nuclear Force

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A closed pipe is a cylinder, tube, or canal of some sort that is closed at one end and open at the other. Sound waves enter through the open end, which is why there can be no sound in a pipe closed on both ends. At the closed end is a node of the standing wave because no motion in the direction the wave is traveling is allowed here. At the open end is the where the maximum amount of motion can occur, and thus this part is where an antinode exists. The length the wave has to travel from one end of this pipe to the other is the distance l (lowercase L, not the number 1).


With a fixed distance for a standing wave to travel through, many different possibilities exist for how many wavelengths can fit from one end of the closed pipe to the other. The most basic of these possibilities is diagrammed above. The first possibility shown is called the "Fundamental Frequency" and is numbered n=1. It shows what the longest wavelength of a sound in this pipe would look like. At the left is the open part of the pipe, which corresponds to an antinode for each case. At the other end is the closed part, which dictates that each situation end in a node at that side. When n=1, a fourth of a wavelength travels through the pipe. The next case is called the "First overtone," the next is called the "Second Overtone" and so on if the diagram were to continue the pattern. For the first overtone, n=2 (simply to signify which situation is happening- if we want to perform some calculations knowing a wave is traveling according to the third overtone, for example, we can plug the value for n, which is 4, into an appropriate equation). When n=2, 3/4 of a wavelength fits into a closed pipe of length l. When n=3, 5/4 of a wavelength fits. See a pattern? With an increase of 1 in the value of n, there is an increase of 2/4 wavelengths. This relationship can be described as l having (2n-1)/4 wavelengths. Solving for lambda gets the result that lambda = 4l/(2n-1).

Let's test this equation out, shall we? What if a closed pipe is 0.5 m long, what is the wavelength of a wave at the fundamental frequency?
Solution: Just plug in l=0.5 m and n=1 into the equation to get that 4(0.5 m)/(2(1)-1) = 2 m.
We can take this one more step, to find the fundamental frequency itself for this particular wave and pipe. Knowing that v=f(lambda), and that the speed of sound is approximately 330 m/s, it is possible to plug in the value for the wavelength obtained before, and the known speed to find the frequency. We get that 330 m/s / 2 m = 165 Hz.



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