Question

Question: (I) (a) What resonant frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep, if you assumed it was a closed tube? (b) How would that change if it was one-third full of soda?

Short answer

The frequency of the empty soda bottle is \(360.4{\rm{ Hz}}\) and (b) The frequency of the one-third full soda bottle is \(540.6{\rm{ Hz}}\).

Step-by-step solution

Step-1:-Fundamental frequency of a closed tube

In order to determine the fundamental frequency, consider the standard value of speed of sound that is 346 m/s.

Step-2:-Given data

The length of the bottle is \(l = 24{\rm{ cm}}\).

The speed of sound is \(\nu = 346{\rm{ m/s}}\).

Step-3:-Calculation of resonant frequency of empty soda bottle

The relation of resonant frequency is given by,

\(f = \frac{\nu }{{4l}}\)

Substitute the values in the above relation.

\(\begin{array}{c}f = \frac{{346{\rm{ m/s}}}}{{4 \times \left( {24\;{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}}\\ = 360.4{\rm{ Hz}}\end{array}\)

The frequency of the empty soda bottle is \(360.4{\rm{ Hz}}\).

Step-4:-Calculation of resonant frequency of one-third full soda bottle

If the bottle is one-third full, the length reduces to \(l' = 16{\rm{ cm}}\).

The relation of resonant frequency is given by,

\(f = \frac{\nu }{{4l'}}\)

Substitute the values in the above relation.

\(\begin{array}{c}f = \frac{{346{\rm{ m/s}}}}{{4 \times \left( {16\;{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}}\\ = 540.6{\rm{ Hz}}\end{array}\)

The frequency of the bottle with one-third full of soda is \(540.6{\rm{ Hz}}\).