Help Solving Physics Problem: Air Column Frequency Overtones

[Tia_Bear]

Hi,

I'm new here, so I don't really know how this works, but if someone could help me solve this physics problem, that would be great!

An air column open at both ends, has a fundamental frequency of 2000 Hz. What are the frequencies of the first two overtones?

I have been stuck on this problem for a while, and can't seem to figure it out! Could someone please explain it to me? Thanks!

 

[learningphysics]

What is the definition of an overtone?

 

[m2003]

I would be happy to help you solve this physics problem. First, let's define some key terms. The fundamental frequency is the lowest frequency at which a system can vibrate, in this case, the air column. Overtones, also known as harmonics, are higher frequencies that are multiples of the fundamental frequency. The first overtone is the second harmonic, which has a frequency twice that of the fundamental frequency. The second overtone is the third harmonic, with a frequency three times that of the fundamental frequency.

To solve this problem, we can use the formula for the frequency of an open pipe: f = nv/2L, where n is the harmonic number, v is the speed of sound in air (approximately 343 m/s at room temperature), and L is the length of the air column. For the fundamental frequency, n=1, so we can rearrange the formula to solve for L: L = nv/2f. Plugging in the given values, we get L = (1)(343 m/s)/(2)(2000 Hz) = 0.086 m.

To find the first overtone, we use the same formula with n=2: f = (2)(343 m/s)/(2)(0.086 m) = 4000 Hz. Similarly, for the second overtone, we use n=3: f = (3)(343 m/s)/(2)(0.086 m) = 6000 Hz.

Therefore, the first overtone has a frequency of 4000 Hz and the second overtone has a frequency of 6000 Hz. I hope this explanation helps you understand the problem and how to solve it. Keep practicing and you will become more familiar with these types of physics problems. Good luck!