Chapter 12, Problem 29P

(a)

To determine

To Find: The distance from the end of the string.

Answer to Problem 29P

The distance from the end of the string.is $0.1\text{8}\text{m}$ .

Explanation of Solution

Given:

Length, ${\mathrm{l}}_{\mathrm{g}}=0.\text{73 m}$

Frequency at point A, ${\mathrm{f}}_{\mathrm{A}} =\text{44}0\text{ Hz}$

Frequency at point E, ${\mathrm{f}}_{\mathrm{E}}=\text{33}0\text{Hz}$ .

Formula used:

The fundamental frequency is given as,

  $\mathrm{f}=\frac{\mathrm{v}}{2\mathrm{L}}$

Where, v is the velocity and L is the length of the string

Calculation:

According to the formula, the frequency is inversely proportional to the length to length. Therefore, the product of frequency and length is constant such that,

  $\mathrm{f}\propto \frac{1}{\mathrm{L}}\Rightarrow \mathrm{f}\mathrm{L}=\text{Const}\text{.}$

So, apply this formula for the given case,

  $\begin{array}{c}{\mathrm{f}}_{\mathrm{E}}{\mathrm{L}}_{\mathrm{E}}={\mathrm{f}}_{\mathrm{A}}{\mathrm{L}}_{\mathrm{A}}\\ {\mathrm{L}}_{\mathrm{A}}={\mathrm{L}}_{\mathrm{E}}\frac{{\mathrm{f}}_{\mathrm{E}}}{{\mathrm{f}}_{\mathrm{A}}}\\ =\left(0.73\text{m}\right)\times \left(\frac{330\text{Hz}}{440\text{Hz}}\right)\\ =0.5475\text{m}\end{array}$

The length of the string from the nut of the guitar is

  $\left(0.73\text{m}\right)-0.5475\text{}\text{m}=0.18\text{m}$

Conclusion:

The string must be fingered $0.18\text{m}$ from the end.

(b)

To determine

To Find: The wavelength on the string.

Answer to Problem 29P

  $1.1\text{m}$

Explanation of Solution

Given:

Frequency at point A, ${\mathrm{f}}_{\mathrm{A}} =\text{44}0\text{ Hz}$

Speed of wave, $\mathrm{v} =\text{49}0.\text{56 m}/\text{s}$ .

Formula used:

The wavelength is calculated as,

  $\mathrm{\lambda }=2\mathrm{L}$

Where L is the length of the string

Calculation:

The string is vibrating in its fundamental mode if it is fixed at the both ends. Therefore, the wavelength is equal to the two times of the length of the string. So, the wavelength is calculated as,

  $\mathrm{\lambda }=2\mathrm{L}=2\times \left(0.5475\text{m}\right)=1.1\text{m}$

Conclusion:

The wavelength of the 440 Hz frequency wave is $1.1\text{m}$ .

(c)

To determine

To Calculate: The frequency and wavelength of the sound wave produced in air.

Answer to Problem 29P

λ= $0.78\text{m}$ and $\mathrm{f}=440\text{Hz}$

Explanation of Solution

Given:

Temperature ( T ) = ${20}^{\text{o}}\text{C}$

Formula used:

The wavelength is given as,

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

Where, $\mathrm{v}$ and $\mathrm{f}$ is the velocity and frequency

Calculation:

The frequency of the string is must be equal to the that of the sound, that is $440\text{Hz}$ .

Now, the wavelength is,

  $\begin{array}{c}\mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}}\\ =\frac{\text{343}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}{\text{440}\text{Hz}}\\ =0.78\text{m}\end{array}$

Conclusion:

The wavelength of the soundwave produced at ${20}^{\text{o}}\text{C}$ is 0 $0.78\text{m}$ and the frequency is 440 Hz.