Chapter 22, Problem 16E

(a)

To determine

The number of harmonics appear below $1350\text{Hz}$.

Answer to Problem 16E

The number of harmonics appear below $1350\text{Hz}$ is $5$.

Explanation of Solution

Write the expression for frequency of column.

  $f_n=\frac{\left(2n-1\right)c}{4l}$        (I)

Here, $c$ is speed of sound, $l$ is the length of clarinet, $n$ is number of harmonics and $f_n$ is frequency of column.

Rearrange the above expression in terms of $l$.

  $l=\frac{\left(2n-1\right)c}{4f_n}$        (II)

Conclusion:

Substitute $147\text{Hz}$ for $f_1$, $1$ for $n$ and $344\text{m}/\text{s}$ for $c$ in equation (II).

  $\begin{array}{c}l=\frac{\left(2\left(1\right)-1\right)\left(344\text{m}/\text{s}\right)}{4\left(147\text{Hz}\right)}\\ =\frac{\left(2-1\right)\left(344\text{m}/\text{s}\right)}{588\text{Hz}}\\ =0.585\text{m}\end{array}$

For first overtone:

Substitute $2$ for $n$, $0.585\text{m}$ for $l$ and $344\text{m}/\text{s}$ for $c$ in equation (I).

  $\begin{array}{c}{f}_2=\frac{\left(2\left(2\right)-1\right)\left(344\text{m}/\text{s}\right)}{4\left(0.585\text{m}\right)}\\ =\frac{\left(4-1\right)\left(344\text{m}/\text{s}\right)}{4\left(0.585\text{m}\right)}\\ =441\text{Hz}\end{array}$

The table for number of frequencies is given below.

Harmonics$\left(n\right)$	Frequency$\left(f_n\right)$
$1$	$147\text{Hz}$
$2$	$441\text{Hz}$
$3$	$735\text{Hz}$
$4$	$1029\text{Hz}$
$5$	$1323\text{Hz}$
$6$	$1617\text{Hz}$

The value of fifth overtone is greater than $1350\text{Hz}$; therefore, the frequency up to fourth overtone would only appear below $1350\text{Hz}$.

Thus, the number of harmonics appear below $1350\text{Hz}$ is $5$.

(b)

To determine

The number of harmonics appear below $1350\text{Hz}$.

Answer to Problem 16E

The number of harmonics appear below $1350\text{Hz}$ is $9$.

Explanation of Solution

Write the expression for frequency of column.

  $f_n=\frac{nc}{2l}$        (III)

Here, $c$ is speed of sound, $l$ is the length of clarinet, $n$ is number of harmonics and $f_n$ is frequency of column.

Rearrange the above expression in terms of $l$.

  $l=\frac{nc}{2f_n}$        (IV)

Conclusion:

Substitute $147\text{Hz}$ for $f_1$, $1$ for $n$ and $344\text{m}/\text{s}$ for $c$ in equation (IV).

  $\begin{array}{c}l=\frac{1\left(344\text{m}/\text{s}\right)}{2\left(147\text{Hz}\right)}\\ =1.170\text{m}\end{array}$

For first overtone:

Substitute $2$ for $n$, $1.170\text{m}$ for $l$ and $344\text{m}/\text{s}$ for $c$ in equation (III).

  $\begin{array}{c}{f}_2=\frac{2\left(344\text{m}/\text{s}\right)}{2\left(1.170\text{m}\right)}\\ =294\text{Hz}\end{array}$

The table for number of frequencies is given below.

Harmonics$\left(n\right)$	Frequency$\left(f_n\right)$
$1$	$147\text{Hz}$
$2$	$294\text{Hz}$
$3$	$441\text{Hz}$
$4$	$588\text{Hz}$
$5$	$735\text{Hz}$
$6$	$882\text{Hz}$
$7$	$1029\text{Hz}$
$8$	$1176\text{Hz}$
$9$	$1323\text{Hz}$
$10$	$1470\text{Hz}$

The value of ninth overtone is greater than $1350\text{Hz}$; therefore, the frequency up to eighth overtone appear below $1350\text{Hz}$.

Thus, the number of harmonics appear below $1350\text{Hz}$ is $9$.