In an organ pipe open at one end closed at the other, the fundamental frequency

Question

In an organ pipe open at one end closed at the other, the fundamental frequency emitted is '210 Hz. (The speed of sound is 340 m/s), Determine the next overtone frequency and the length of the pipe. A referee s whistle sounds at '2000 Hz. Since you know the speed of sound is 340 m/s, find the apparent frequency if you're moving at 40 m/s toward the whistle and away from the whistle. A railroad crossing warning signal sounds at 120 Hz. If you are approaching the crossing at 3% of the speed of sound, find the frequency you would hear. If you were going away from the crossing at 3% of sound speed, find the observed frequency.

Explanation / Answer

It is so much easier to draw these for a good explanation, but here goes.

For an open/closed pipe, when we draw the harmonic the fundamental has a node at the closed end and an antinode at the open end with nothing in between. In terms of wavelength, from a node to the first antinode is 1/4 lambda. So the length of the pipe is equal to lambda/4:

L = lambda / 4

There is a simple relationship between wavelength, frequency and wave speed:

(lambda) * ( f ) = v

The speed of sound in air is typically given as 343 m/s and the frequency is 240 Hz. Plug these into get: lambda = 343m/s / 240Hz = 1.429m and thus
L = 1.429 / 4 = 0.357m


If you draw the next overtone, it has a node at closed end, an 2nd node in the middle of the pipe and then an anti node at the open end. This is not quite 1 full wavelength, but instead is 3/4 lambda. So for the first overtone, L = 3/4 lambda.

We have the length of the pipe, so plugging in we get that the wavelength of the first overtone is:
lambda = 4/3 * 0.357 = 0.476m

Plug this into the frequency-wavelength relation to get:
(0.476) f = 343 = 720 Hz

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.