Physics Help Forum Why does resonant frequency become lower as water is added to a bottle?

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 Sep 9th 2019, 02:38 AM #1 Junior Member   Join Date: Sep 2019 Posts: 4 Why does resonant frequency become lower as water is added to a bottle? Hello! I am an IB student currently working on my Internal Assessment, which is an investigation on the relationship between the volume of water in a glass bottle and the frequency of the sound made when the bottle is hit with a metal knife. I have collected my data, and found that the frequency indeed became lower as more water was added to the bottle. This supports my hypothesis. What I'm having trouble understanding, however, is the reason this happens. From my research I found this answer on the Physics Stackexchange Forum, which essentially states the following: The bottle acts as an air column for the vibrations in the air. Changing the volume of water changes the length of this air column. Therefore it will contain a gradually changing magnitude of the wave's wavelength. This, through the formula of f=v/λ, affects the frequency. I don't really understand though whether this is actually saying the frequency becomes higher or lower when the volume of water is increased. Furthermore, in my case, is it actually the air that is vibrating, or the glass? What would this mean for my hypothesis? I know that the hypothesis is correct, because my data clearly shows a downward relationship between volume of water and resonant frequency. I simply can not find what the reasoning behind this is. I know that this is probably a lot to ask, but I would really appreciate some help on this, as I am determined to do well in this assessment and it contributes greatly to my final grade. Thank you!
 Sep 9th 2019, 04:39 PM #2 Junior Member     Join Date: Jan 2019 Posts: 20 Increase in water level decreases the length, L, of the half-open air column which in turn decreases the wavelength of the fundamental resonant frequency. With $v$ remaining constant, $\dfrac{v}{\color{red}{\downarrow} \lambda} = \color{red}{\uparrow} f$ Attached Thumbnails
 Sep 13th 2019, 12:39 AM #3 Junior Member   Join Date: Sep 2019 Posts: 4 Thanks! That certainly helped me understand the logic behind the process; however it also showed me that I seem to be using the wrong formula as my case is not a perfect cylinder, but a bottle, and the recorded frequency actually went down as I added water to the bottle. I believe the formula I need to be using is the one for cavity resonance: The problem is that this photo says "larger volume gives lower frequency". Yet in my case, lowering the volume (of air) gave a lower frequency. Would you mind helping me out with this? Edit: Well, I've done more research and all I keep finding is formulas for frequency when blowing air into the mouth of the bottle, or something along those lines. Does anyone know if there is a formula to use when the sound is produced by hitting the bottle, instead of blowing air into it? Last edited by Graystripe; Sep 13th 2019 at 12:52 AM.
 Sep 13th 2019, 08:58 AM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,351 It does seem that most of the literature regarding experiments with bottles filled with water has to do with the frequency of sound when you blow across the open mouth of the bottle. In that case the tone you hear is due to the length of the air column - as you add water the column of air gets shorter and the frequency of the sound you make goes up. But in your case what you are hearing is the vibration of the glass, not the air inside the bottle. As you add water you dampen the vibrations of the wall of the bottle, which serves to slow the movement of the bottle wall, which leads to a lower frequency of vibrations of the bottle wall. If you like I can show you the differential equation of a vibration system with dampening, which shows the the fundamental frequency of vibration is reduced as damping is increased, but it might be a little advanced for your purposes.
 Sep 13th 2019, 10:32 AM #5 Junior Member   Join Date: Sep 2019 Posts: 4 Thanks a lot for the clarification. The equation would be great, as would any other information or links you might have regarding the dampening of the vibrations. It may be too advanced, but I'd at least like to try and understand the physics behind my results; I know that this is asking for quite a lot, and of course I don't want you guys to feel like you're doing all my work for me, but I have to admit I'm a bit stumped here and would really appreciate the help if you have anything else.
 Sep 13th 2019, 11:14 AM #6 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,351 Here's a reference to a site that works through the mathematics of a damped system consisting of an ideal spring, damper, and mass: https://www.brown.edu/Departments/En...ree_damped.htm The key to understanding the math is to understand that the system consists of a mass that is vibrating and restrained by a spring, which provides a force proportional and opposite to the displacement of the mass, plus a damper (or shock absorber, if you're familiar with how a car's suspension works) which provides a force proportional and opposite to the velocity of the mass. Now a bottle is not exactly the same as an ideal spring, mass and damper, but the ideas are consistent. The natural frequency of a damped systems is lower than the natural frequency of an undamped system consisting of the same mass and spring. Specifically: $\displaystyle \omega_d = \omega_n \sqrt {1 - \zeta^2}$ where $\omega_d$ is the frequency of the damped system, $\omega_n$ is the natural frequency of an undamped system, and $\zeta$ is the damping coefficient. For purposes here we can say that zeta takes on a value somewhere between 0 and 1 (i.e. the system is underdamped, as it must be, or otherwise you wouldn't hear a tone when you strike the bottle).
 Today, 02:35 AM #7 Junior Member   Join Date: Sep 2019 Posts: 4 Thank you, this is starting to make a lot more sense now. So would increasing the volume of water in the bottle increase the damping coefficient, hence lowering the frequency of the damped system? And the natural frequency remains constant, i.e. the frequency of the bottle with no water added? Secondly, do we actually know what the damping coefficient is for my experiment or is that far too complicated to figure out without doing the experiment and just using the other two variables to calculate it? The reason I am asking this is because if there was some simple way to find out ζ, I could use that value to calculate theoretical results for the damped frequencies and compare those to the actual results. If not though, I'm sure it'll be fine, I'd just like to know whether that's at all possible. Even writing about why I wasn't able to find out zeta before conducting the experiment would help me, I believe. Edit: Oh, and one more thing if you don't mind; is the frequency I am measuring technically called the resonant frequency, the fundamental frequency, or are both terms fine? Last edited by Graystripe; Today at 02:43 AM.

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