Chapter 15.2, Problem 16PP

(a)

To determine

To Find: The lowest frequency that is resonant for a bugle.

Answer to Problem 16PP

  $64.72\text{Hz}$

Explanation of Solution

Given: Length of the bugle, $L=2.65\text{m}$

Speed of the sound, $v=343\text{m/s}$

Formula used:

Frequency, $f=\frac{v}{\lambda }$

Where,

  $v$ is the velocity

  $\lambda$ is the wavelength

Calculation:

For open pipe,

  $\begin{array}{l}L=\frac{\lambda }{2}\\ \lambda =2L\\ \lambda =2\left(2.65\right)\\ \lambda =5.3\text{m}\end{array}$

Now, frequency can be calculated as:

  $\begin{array}{l}f=\frac{343}{5.30}\\ f=64.72\text{Hz}\end{array}$

Conclusion:

  $f=64.72\text{Hz}$ is the lowest frequency that is resonant for a bugle.

(a)

To determine

To Find: The next two resonant frequencies for a bugle.

Answer to Problem 16PP

  $129.43\text{Hz, 194}\text{.15}\text{Hz}$

Explanation of Solution

Given:

Length of the bugle, $L=2.65\text{m}$

Speed of the sound, $v=343\text{m/s}$

Formula used:

Frequency, $f=\frac{v}{\lambda }$

Where,

  $v$ is the velocity

  $\lambda$ is the wavelength

Calculation:

Next two resonant frequency will take place at $\frac{v}{L}$ and $\frac{3v}{2L}$

  $\begin{array}{l}{f}_1=\frac{v}{L}\\ f_1=\frac{343}{2.65}=129.43\text{Hz}\end{array}$

And

  $\begin{array}{l}{f}_2=\frac{3v}{2L}\\ f_2=\frac{3\times 343}{2\times 2.65}=194.15\text{Hz}\end{array}$

Conclusion:

  $129.43\text{Hz, 194}\text{.15}\text{Hz}$ are the next two resonant frequencies for a bugle.