Chapter 17, Problem 24P

To determine

The frequencies, which would sound the richest because of resonance.

Explanation of Solution

Given info: The dimension of the shower stall is $86.0\text{cm}\times 86.0\text{cm}\times 210\text{cm}$. The range of frequencies of the voices lies from $130\text{Hz}$ to $2000\text{Hz}$. The speed of the sound in the hot air is $355\text{m}/\text{s}$.

Expression for the fundamental or first harmonic frequency is,

$f_1=\frac{v}{4L}$

Here,

$v$ is the velocity of the sound in hot air.

$L$ is the length of the shower stall.

Substitute $355\text{m}/\text{s}$ for $v$ and $210\text{cm}$ for $L$ in the above equation.

$\begin{array}{c}{f}_1=\frac{355\text{m}/\text{s}}{4\times \left(210\text{cm}\right)}\\ =\frac{355\text{m}/\text{s}}{4\times \left(210\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =\frac{355\text{m}/\text{s}}{4\times \left(2.10\text{m}\right)}\\ =42.262\text{Hz}\end{array}$

Expression for the third harmonic frequency is,

$f_3=\frac{3v}{4L}$

Substitute $355\text{m}/\text{s}$ for $v$ and $210\text{cm}$ for $L$ in the above equation.

$\begin{array}{c}{f}_3=\frac{3\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\right)}\\ =\frac{3\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =\frac{3\times \left(355\text{m}/\text{s}\right)}{4\times \left(2.10\text{m}\right)}\\ =126.786\text{Hz}\end{array}$

Expression for the fifth harmonic frequency is,

$f_5=\frac{5v}{4L}$

Substitute $355\text{m}/\text{s}$ for $v$ and $210\text{cm}$ for $L$ in the above equation.

$\begin{array}{c}{f}_5=\frac{5\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\right)}\\ =\frac{5\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =\frac{5\times \left(355\text{m}/\text{s}\right)}{4\times \left(2.10\text{m}\right)}\\ =211.31\text{Hz}\end{array}$

Expression for the seventh harmonic frequency is,

$f_7=\frac{7v}{4L}$

Substitute $355\text{m}/\text{s}$ for $v$ and $210\text{cm}$ for $L$ in the above equation.

$\begin{array}{c}{f}_7=\frac{7\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\right)}\\ =\frac{7\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =\frac{7\times \left(355\text{m}/\text{s}\right)}{4\times \left(2.10\text{m}\right)}\\ =295.834\text{Hz}\end{array}$

Expression for the ninth harmonic frequency is,

$f_9=\frac{9v}{4L}$

Substitute $355\text{m}/\text{s}$ for $v$ and $210\text{cm}$ for $L$ in the above equation.

$\begin{array}{c}{f}_9=\frac{9\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\right)}\\ =\frac{9\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =\frac{9\times \left(355\text{m}/\text{s}\right)}{4\times \left(2.10\text{m}\right)}\\ =380.358\text{Hz}\end{array}$

Expression for the eleventh harmonic frequency is,

$f_{11}=\frac{11v}{4L}$

Substitute $355\text{m}/\text{s}$ for $v$ and $210\text{cm}$ for $L$ in the above equation.

$\begin{array}{c}{f}_{11}=\frac{11\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\right)}\\ =\frac{11\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =\frac{11\times \left(355\text{m}/\text{s}\right)}{4\times \left(2.10\text{m}\right)}\\ =464.881\text{Hz}\end{array}$

Similarly, for maximum resonance frequency,

$f_{\text{n}}=\frac{\left(2n+1\right)v}{4L}$

Substitute $2000\text{Hz}$ for $f_{\text{n}}$, $355\text{m}/\text{s}$ for $v$ and $210\text{cm}$ for $L$ in the above equation.

$\begin{array}{c}2000\text{Hz}=\frac{\left(2n+1\right)\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\right)}\\ 2000\text{Hz}=\frac{\left(2n+1\right)\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =\frac{\left(2n+1\right)\times \left(355\text{m}/\text{s}\right)}{4\times \left(210\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\end{array}$

Simplify further,

$\begin{array}{c}2000\text{Hz}=\frac{\left(2n+1\right)\times \left(355\text{m}/\text{s}\right)}{4\times \left(2.10\text{m}\right)}\\ n\approx 47\end{array}$

For $n=47$,

$\begin{array}{c}{f}_{47}=47\times \left(42.262\text{Hz}\right)\\ =1986.314\text{Hz}\end{array}$

Conclusion:

Therefore, the frequencies, which would sound the richest because of resonance, are $211.31\text{Hz}$, $295.834\text{Hz}$, $380.358\text{Hz}$, $464.881\text{Hz}$ up to $1986.314\text{Hz}$.