Chapter 12, Problem 37P

(a)

To determine

To calculate: The number of overtones present within the audible range.

Answer to Problem 37P

  $n=248$ .

Explanation of Solution

Given:

Length of the tube $L=2.40\text{ m}$

Temperature $T=20°\text{C}$

Formula used:

The frequency is given as,

  $f=\frac{nv}{2L}$

Where, $L$ is the string length and $v$ is the speed of the sound

Calculation:

For the audible, the frequency must be less than $\text{20}\text{kHz}$ . Therefore, the harmonics for open pipe are given as,

  $\begin{array}{l}\frac{nv}{2L}<2\times {10}^4\text{Hz}\\ \Rightarrow n<\frac{2L}{v}\times 2\times {10}^4\text{Hz}\\ n<249.6\end{array}$

The number of harmonics is $249.6$ and so, the number of overtones is $248$ .

Conclusion:

The number of overtones present within the audible range for open pipe is $248$

(b)

To determine

To Find:The number of overtones are present within the audible range

Answer to Problem 37P

  $n=249$ overtones are present.

Explanation of Solution

Given:

Length of the tube $L=2.40\text{ m}$

Temperature $T=20°\text{C}$

Formula used:

The harmonics for the closed pipe is calculated for the odd values of n that is given as,

  $f=\frac{nv}{2L}$

Where, $L$ is the string length and $v$ is the speed of the sound

Calculation:

The harmonics for the closed pipe is calculated for the odd values of n. For the audible, the frequency must be less than $\text{20}\text{kHz}$ also. Therefore, the harmonics for closed pipe are given as,

  $\begin{array}{l}\frac{nv}{2L}<2\times {10}^4\text{Hz}\\ \Rightarrow \\ n<\frac{\text{2×2}\text{.14m}}{\text{343}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}{\text{×2×10}}^{\text{4}}\text{Hz}\\ n<499.1\end{array}$

But the values of n must be odd values such that $n=1,3,\mathrm{5.......499}$

Conclusion:

Hence, there are $250$ harmonics and the number of overtones is $249$