Chapter 15, Problem 61A

(a)

To determine

To calculate: The speed of the wave on the string.

Answer to Problem 61A

The speed of the wave is $\underset{\_}{254.8\text{ m/s}}$ .

Explanation of Solution

Given:

Length of guitar string = $65\text{ cm}$ .

Lowest frequency = $\text{196 Hz}$

Formula used:

Wavelength is given by the formula,

$\lambda =2L$

Where $L$ is length.

Speed of the wave is given by the formula,

$v=f\lambda$

Where $f$ is frequency and $\lambda$ is wavelength.

Calculation:

Wavelength of the air column is calculated as:

$\begin{array}{l}\lambda =2L\\ =2\times 65\text{ cm}\\ =130\text{ cm}\\ =1.3\text{ m}\end{array}$

The speed of the wave is calculated as:

$\begin{array}{l}v=f\lambda \\ =196\text{ Hz}\times 1.3\text{ m}\\ =254.8\text{ m/s}\end{array}$

Conclusion:

The speed of the wave is $\underset{\_}{254.8\text{ m/s}}$

(b)

To determine

To calculate: The next two higher resonating frequencies of the string.

Answer to Problem 61A

The second higher frequency is $\underset{\_}{392\text{ Hz}}$ .

The third higher frequency is $\underset{\_}{\text{588 Hz}}$ .

Explanation of Solution

Given:

Length of guitar string = $65\text{ cm}$ .

Lowest frequency = $\text{196 Hz}$

Formula used:

The second wavelength is given by the formula,

${\lambda }_2=L$

Where $L$ is length.

The third wavelength is given by the formula,

${\lambda }_3=\frac{2L}{3}$

Where $L$ is length.

Frequency is given by the formula,

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

Where $v$ is velocity and $\lambda$ is wavelength.

Calculation:

Second wavelength of the air column is calculated as:

$\begin{array}{l}{\lambda }_2=L\\ =65\text{ cm}\\ =0.65\text{ m}\end{array}$

Second higher frequency is calculated as:

  $\begin{array}{l}{f}_2=\frac{v}{{\lambda }_2}\\ \text{ =}\frac{254.8\text{ m/s}}{0.65\text{ m}}\\ \text{ =}392\text{ Hz}\end{array}$

Third wavelength of the air column is calculated as:

$\begin{array}{l}{\lambda }_3=\frac{2L}{3}\\ =\frac{2\times 65\text{ cm}}{3}\\ =43.\overline{33}\text{ cm}\\ \text{ =0}\text{.4}\overline{\text{33}}\text{ m}\end{array}$

Third higher frequency is calculated as:

  $\begin{array}{l}{f}_3=\frac{v}{{\lambda }_3}\\ \text{ =}\frac{254.8\text{ m/s}}{0.4\overline{33}\text{ m}}\\ \text{ =588 Hz}\end{array}$

Conclusion:

The second higher frequency is $\underset{\_}{392\text{ Hz}}$ .

The third higher frequency is $\underset{\_}{\text{588 Hz}}$ .