Chapter 19, Problem 79PQ

(a)

To determine

The mass per unit length of the guitar string.

Answer to Problem 79PQ

The mass per unit length of the guitar string is $\underset{\_}{3.95\times {10}^{-4}\text{kg}/\text{m}}$.

Explanation of Solution

Write the expression for volume of the wire.

$V=LA$ (I)

Here, $V$ is the volume of the wire, $L$ is the length of the wire, and $A$ is the cross sectional area of the wire.

The wire has a circular cross section.

Write the expression for cross sectional area of wire.

$A=\pi r^2$ (II)

Here, $r$ is the radius of the wire.

Use equation (II) in (I).

$V=L\left(\pi r^2\right)$ (III)

Write the expression for linear mass density of the wire.

$\mu =\frac{m}{L}$ (IV)

Here, $\mu$ is the linear mass density, and $m$ is the mass of the guitar string.

Rewrite equation (I) to find $L$.

$L=\frac{V}{A}$

Use above equation in (IV).

$\mu =A\frac{m}{V}$ (V)

Use equation (II) in (V).

$\mu =\pi r^2\frac{m}{V}$ (VI)

Write the expression for density of the wire.

$\rho =\frac{m}{V}$ (VII)

Here, $\rho$ is the density of the wire.

Use equation (VII) in (VI).

$\mu =\pi r^2\rho$ . (VIII)

Write the expression to find radius of the wire.

$r=\frac{d}{2}$ (IX)

Here, $d$ is the diameter of the wire.

Conclusion:

Substitute $0.0254\text{cm}$ for $d$ in equation (IX) to find $r$.

  $\begin{array}{c}r=\frac{0.0254\text{cm}\times \frac{1\text{m}}{100\text{cm}}}{2}\\ =1.27\times {10}^{-4}\text{m}\end{array}$

Substitute $1.27\times {10}^{-4}\text{m}$ for $r$, and $7.80\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in equation (VIII) to find $\mu$.

  $\begin{array}{c}\mu =\pi {\left(1.27\times {10}^{-4}\text{m}\right)}^2\left(7.80\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\\ =3.95\times {10}^{-4}\text{kg}/\text{m}\end{array}$

Therefore, the mass per unit length of the guitar string is $\underset{\_}{3.95\times {10}^{-4}\text{kg}/\text{m}}$.

(b)

To determine

The tension in the guitar string.

Answer to Problem 79PQ

The tension in the guitar string is $\underset{\_}{26.3\text{N}}$.

Explanation of Solution

Write the expression for the fundamental frequency of vibration of the guitar string.

$f_1=\frac{v}{2L}$ (X)

Here, $f_1$ is the fundamental frequency of vibration of the string, and $v$ is the speed of the waves.

Write the expression to find the speed of the waves in terms of the tension and mass density of the wire.

$v=\sqrt{\frac{F_T}{\mu }}$

Here, $F_T$ is the force of tension in the wire.

Use above expression in equation (X).

$f_1=\frac{1}{2L}\sqrt{\frac{F_T}{\mu }}$ (XI)

Solve equation (XI) for $F_T$.

$F_T=\mu {\left(2L{f}_1\right)}^2$ (XII)

Conclusion:

Substitute $3.95\times {10}^{-4}\text{kg}/\text{m}$ for $\mu$, $65.78\text{cm}$ for $L$, and $196\text{Hz}$ for $f_1$ in equation (XII) to find $F_T$.

  $\begin{array}{c}{F}_T=\left(3.95\times {10}^{-4}\text{kg}/\text{m}\right){\left[2\left(65.78\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)196\text{Hz}\right]}^2\\ =26.3\text{N}\end{array}$

Therefore, the tension in the guitar string is $\underset{\_}{26.3\text{N}}$.

(c)

To determine

The tension in the guitar string if the temperature is increased to $45.0°\text{C}$.

Answer to Problem 79PQ

The tension in the guitar string if the temperature is increased to $45.0°\text{C}$ is $\underset{\_}{23.5\text{N}}$.

Explanation of Solution

Write the expression to find the unstressed length of the guitar string at $20.0°\text{C}$.

$L=L_{\text{natural}}\left(1+\frac{F_T}{AY}\right)$ (XIII)

Here, $L$ is the scale length of the guitar, $L_{\text{natural}}$ is the unstressed length of the guitar string at $20.0°\text{C}$, $Y$ is the Young’s Modulus of the material of the guitar string.

Solve equation (XIII) to find $L_{\text{natural}}$.

$L_{\text{natural}}=\frac{L}{1+F_T/AY}$ (XIV)

Write the expression to find the unstressed length at $45.0°\text{C}$.

$L_{45°\text{C}}=L_{\text{natural}}\left[1+\alpha \left(T_f-F_i\right)\right]$ (XV)

Here, $L_{45°\text{C}}$ is the unstressed length of wire at $45.0°\text{C}$, $\alpha$ linear expansion coefficient, $T_f$ is the final temperature, and $T_i$ is the initial temperature.

Write the equation to find the scale length of the string.

$L=L_{45°\text{C}}\left[1+\frac{{F^{\prime }}_T}{AY}\right]$ (XVI)

Here, ${F^{\prime }}_T$ is the force of tension at $45.0°\text{C}$.

Solve equation (XVI) for ${F^{\prime }}_T$.

${F^{\prime }}_T=AY\left[\frac{L}{L_{45°\text{C}}}-1\right]$ (XVII)

Conclusion:

Substitute $1.27\times {10}^{-4}\text{m}$ for $r$ in equation (II) to find $A$.

  $\begin{array}{c}A=\pi {\left(1.27\times {10}^{-4}\text{m}\right)}^2\\ =5.06\times {10}^{-8}{\text{m}}^2\end{array}$

Substitute $26.3\text{N}$ for $F_T$, $65.78\text{cm}$ for $L$, $5.06\times {10}^{-8}{\text{m}}^2$ for $A$, and $20.0\times {10}^{10}\text{Pa}$ for $Y$ in equation (XIV) to find $L_{\text{natural}}$.

  $\begin{array}{c}{L}_{\text{natural}}=\frac{65.78\text{cm}\times \frac{1\text{m}}{100\text{cm}}}{1+\frac{26.3\text{N}}{\left(5.06\times {10}^{-8}{\text{m}}^2\right)\left(20.0\times {10}^{10}\text{Pa}\right)}}\\ =0.6561\text{m}\end{array}$

Substitute $0.6561\text{m}$ for $L_{\text{natural}}$, $11.0\times {10}^{-6}°{\text{C}}^{-1}$ for $\alpha$, $45.0°\text{C}$ for $T_f$, and $20.0°\text{C}$ for $T_i$ in equation (XV) to find $L_{45°\text{C}}$.

  $\begin{array}{c}{L}_{45°\text{C}}=\left(0.6561\text{m}\right)\left[1+\left(11.0\times {10}^{-6}°{\text{C}}^{-1}\right)\left(45.0°\text{C}-20.0°\text{C}\right)\right]\\ =0.6563\text{m}\end{array}$

Substitute $5.06\times {10}^{-8}{\text{m}}^2$ for $A$, $20.0\times {10}^{10}\text{Pa}$ for $Y$, $0.6578\text{m}$ for $L$, and $0.6563\text{m}$ for $L_{45°\text{C}}$ in equation (XVII) to find ${F^{\prime }}_T$.

  $\begin{array}{c}{F^{\prime }}_T=\left(5.06\times {10}^{-8}{\text{m}}^2\right)\left(20.0\times {10}^{10}\text{Pa}\right)\left[\frac{0.6578\text{m}}{0.6563\text{m}}-1\right]\\ =23.5\text{N}\end{array}$

Therefore, the tension in the guitar string if the temperature is increased to $45.0°\text{C}$ is $\underset{\_}{23.5\text{N}}$.

(d)

To determine

The fundamental frequency of guitar string at $45.0°\text{C}$.

Answer to Problem 79PQ

The fundamental frequency of guitar string at $45.0°\text{C}$ is $\underset{\_}{185\text{Hz}}$.

Explanation of Solution

Write the expression for the ratio of frequency at $45.0°\text{C}$ and $20.0°\text{C}$.

$\frac{{f^{\prime }}_1}{f_1}=\sqrt{\frac{{F^{\prime }}_T}{F_T}}$ (XVIII)

Here, ${f^{\prime }}_1$ is the frequency at $45.0°\text{C}$, and ${F^{\prime }}_T$ is the tension at $45.0°\text{C}$.

Solve equation (XVIII) for ${f^{\prime }}_1$.

$f^{\prime }=f_1\sqrt{\frac{{F^{\prime }}_T}{F_T}}$ (XIX)

Conclusion:

Substitute $196\text{Hz}$ for $f_1$, $23.5\text{N}$ for ${F^{\prime }}_T$, and $26.3\text{N}$ for $F_T$ in equation (XIX) to find ${f^{\prime }}_1$.

  $\begin{array}{c}{f^{\prime }}_1=\left(196\text{Hz}\right)\sqrt{\frac{23.5\text{N}}{26.3\text{N}}}\\ =185\text{Hz}\end{array}$

Therefore, the fundamental frequency of guitar string at $45.0°\text{C}$ is $\underset{\_}{185\text{Hz}}$.