Question

Determine the fundamental and first overtone frequencies when you are in a \({\bf{9}}{\bf{.0 - m}}\) long hallway with all doors closed. Model the hallway as a tube closed at both ends.

Short answer

The fundamental frequency of first harmonics is \(19\;{\rm{Hz}}\) and the first overtone frequency of second harmonics is \(38.1\;{\rm{Hz}}\).

Step-by-step solution

Concept of fundamental frequency

When the tube is closed at both ends, it is known as vibrating string fixed at both ends.

The formula for the fundamental frequency which is closed as both ends is given as,

\({f_n} = \frac{{nv}}{{2L}}\).

Given information

Given data:

The length of hallway is \(L = 9\;{\rm{m}}\).

Determination of fundamental frequency for first harmonics

The expression for the fundamental frequency it is given as below for first harmonics:

\({f_n} = \frac{{nv}}{{2L}}\)

Here, \(n\) is the number of harmonics and here it is 1, and \(v\) is the speed of the sound.

Substitute the values in the above equation,

\(\begin{array}{c}{f_1} = \frac{{1 \times 343\;{\rm{m/s}}}}{{2 \times 9\;{\rm{m}}}}\\ = 19\;{\rm{Hz}}\end{array}\)

Thus, the fundamental frequency of first harmonics is \(19\;{\rm{Hz}}\).

Determination of the first overtone frequency of second harmonic

The expression for the first overtone frequency is given as,

\({f_n} = \frac{{nv}}{{2L}}\)

Here, \(n\) is the number harmonics and here it is second harmonics.

Substitute the values in the above equation,

\(\begin{array}{c}{f_2} = \frac{{2 \times 343\;{\rm{m/s}}}}{{2 \times 9\;{\rm{m}}}}\\ = 38.1\;{\rm{Hz}}\end{array}\)

Thus, the first overtone frequency of second harmonics is \(38.1\;{\rm{Hz}}\).