Calculating Resonant Frequency of a Soda Bottle

[Malicious]

Homework Statement

(a)What resonant frequency would you expect from blowing across the top of an empty soda bottle that is 15cm deep?

(b) How would that change if it was one-third of soda

Homework Equations

/\=v/f

The Attempt at a Solution

15cm=1.5m
/\=2L=2(1.5m)=3.0m

(a) After I find /\, I try to solve for frequence (f) by plugging /\ into /\=v/f but I do not have v or f. So, I don't know what to do.

(b) I think the answer would be (1/3)/\ .

[ EDIT ] Nevermind, I figured it out.

 

[anathema]

The resonant frequency would be the same for both the empty and one-third full soda bottle, as it is dependent on the length of the bottle and not the amount of liquid inside. So, the resonant frequency for both would be 3.0m. However, the pitch of the sound produced may be slightly different due to the different densities of air and liquid.

 

[m2003]

The resonant frequency of a soda bottle can be calculated using the equation /\=v/f, where /\ is the wavelength, v is the speed of sound, and f is the frequency. In this case, we are given the depth of the soda bottle (15cm or 0.15m) and asked to find the resonant frequency.

To find v, we can use the equation v=fl, where l is the speed of sound in air (343 m/s). Plugging in the values, we get v=343 m/s x 0.15m = 51.45 m/s.

Now, we can plug in the values for /\ and v into the equation /\=v/f to solve for f. This gives us f=v/\=51.45 m/s / 3.0m = 17.15 Hz.

For part (b), if the soda bottle is one-third full, the depth would be 5cm or 0.05m. Following the same steps as above, we get f=v/\=343 m/s / 0.05m = 6860 Hz. This is a much higher resonant frequency compared to the empty bottle, which makes sense since the depth is now much smaller.