Chapter 19, Problem 19.73CP

Review. A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Find the mass per unit length of this siring. (Use the value 7.86 × 10³ kg/m⁴ for the density.) (b) The fundamental frequency of transverse oscillations of the string is 200 Hz. What is the tension in the string? Next, the temperature is raised to 30.0°C. Find the resulting values of (c) the tension and (d) the fundamental frequency. Assume both the Young’s modulus of 20.0 × 10¹⁰ N/m² and the average coefficient of expansion α = 11.0 × 10^-6 (°C)^-1 have constant values between 0.0°C and 30.0°C.

To determine

The mass per unit length of the string.

Answer to Problem 19.73CP

The mass per unit length of the string is $6.17\times {10}^{-3}\text{Kg}/\text{m}$.

Explanation of Solution

Given Info: The diameter of the string of the steel guitar is $1.00\text{mm}$, the string is stretched between the supports $80.0\text{cm}$ apart, the initial temperature is $0.0°\text{C}$, the density of the steel string is $7.86\times {10}^3\text{Kg}/{\text{m}}^{\text{3}}$.

Formula to calculate the radius of the wire is,

$r=\frac{d}{2}$

Here,

$d$ is the diameter of the wire.

Substitute $2.000\text{mm}$ for $d$ in the above expression.,

$\begin{array}{c}r=\frac{1.000\text{mm}}{2}\left(\frac{{10}^{-3}}{1\text{mm}}\right)\\ =0.5\times {10}^{-3}\text{m}\end{array}$

Thus, the value of the radius is $0.5\times {10}^{-3}\text{m}$.

The area of cross section of the steel string is,

$A=\pi r^2$ (1)

Substitute $0.5\times {10}^{-3}\text{m}$ fort $r$ in the above expression.

$\begin{array}{c}A=\pi {\left(0.5\times {10}^{-3}\text{m}\right)}^2\\ =2.5\pi \times {10}^{-7}{\text{m}}^{\text{2}}\end{array}$

Thus, the area of cross section of the steel string is $2.5\pi \times {10}^{-7}{\text{m}}^{\text{2}}$.

Formula to calculate the mass per unit length of the steel string is,

$\mu =A\rho$

Here,

$\rho$ is the density of the steel.

Substitute $2.5\pi \times {10}^{-7}{\text{m}}^{\text{2}}$ for $A$ and $7.86\times {10}^3\text{Kg}/{\text{m}}^{\text{3}}$ for $\rho$ in the above expression.

$\begin{array}{c}\mu =\left(2.5\pi \times {10}^{-7}{\text{m}}^{\text{2}}\right)\left(7.86\times {10}^3\text{Kg}/{\text{m}}^{\text{3}}\right)\\ =6.17\times {10}^{-3}\text{Kg}/\text{m}\end{array}$

Conclusion:

Therefore, the mass per unit length of the string is $6.17\times {10}^{-3}\text{Kg}/\text{m}$.

To determine

The Tension in the string.

Answer to Problem 19.73CP

The Tension in the string is $632\text{N}$.

Explanation of Solution

Given Info: The diameter of the string of the steel guitar is $1.00\text{mm}$, the string is stretched between the supports $80.0\text{cm}$ apart, the initial temperature is $0.0°\text{C}$, the density of the steel string is $7.86\times {10}^3\text{Kg}/{\text{m}}^{\text{3}}$, the fundamental frequency of the transverse oscillations of the string is $200\text{Hz}$, the temperature is raised to $30.0°\text{C}$, the average coefficient of linear expansion of the steel is $11\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$, the young’s modulus is $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$.

Formula to calculate the fundamental frequency is,

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

Here,

$l$ is the length of the string stretched between the supports.

Rearrange the above expression for $T$.

$T={\left(2lf\right)}^2\mu$

Substitute $80.0\text{cm}$ for $l$, $200\text{Hz}$ for $f$ and $6.17\times {10}^{-3}\text{Kg}/\text{m}$ for $\mu$ in the above expression.

$\begin{array}{c}T={\left(2\left(80.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\left(200\text{Hz}\right)\right)}^2\left(6.17\times {10}^{-3}\text{Kg}/\text{m}\right)\\ =632\text{N}\end{array}$

Conclusion:

Therefore, the Tension in the string is $632\text{N}$.

To determine

The Tension in the string when the temperature is raised to $30.0°\text{C}$.

Answer to Problem 19.73CP

The Tension in the string when the temperature is raised to $30.0°\text{C}$.

is $580\text{N}$.

Explanation of Solution

Given Info: The diameter of the string of the steel guitar is $1.00\text{mm}$, the string is stretched between the supports $80.0\text{cm}$ apart, the initial temperature is $0.0°\text{C}$, the density of the steel string is $7.86\times {10}^3\text{Kg}/{\text{m}}^{\text{3}}$, the fundamental frequency of the transverse oscillations of the string is $200\text{Hz}$, the temperature is raised to $30.0°\text{C}$, the average coefficient of linear expansion of the steel is $11\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$, the young’s modulus is $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$.

Formula for the change in the length, when temperature varies is,

$L_2=l\left(1+\alpha \left(T_{\text{2}}-T_1\right)\right)$ (1)

Here,

$\alpha$ is the coefficient of linear expansion for the brass pendulum.

$T_1$ is the initial temperature of the brass pendulum.

$T_2$ is the final temperature of the brass pendulum.

Substitute $11\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$ for $\alpha$, $80\text{cm}$ for $l$, $0.0°\text{C}$ for $T_1$, $30.0°\text{C}$ for $T_2$ in the above expression.

$\begin{array}{c}{L}_2=\left[\begin{array}{l}\left(80.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ \left(1+11\times {10}^{-6}{\left(°\text{C}\right)}^{-1}\left(30.0°\text{C}-0.0°\text{C}\right)\right)\end{array}\right]\\ =0.80026\text{m}\end{array}$

Thus, the final length of the brass pendulum is $0.80026\text{m}$.

Formula to calculate the tension in the wire is,

$T=Y\left(\frac{L_2-l}{l}\right)A$

Substitute $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$ for $Y$, $0.80026\text{m}$ for $L_2$, $2.5\pi \times {10}^{-7}{\text{m}}^{\text{2}}$ for $A$ in the above expression.

$\begin{array}{c}T=\left(20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}\right)\left(\frac{0.80026\text{m}-0.8\text{m}}{0.8\text{m}}\right)\left(2.5\pi \times {10}^{-7}{\text{m}}^{\text{2}}\right)\\ =580\text{N}\end{array}$

Conclusion:

Therefore, the Tension in the string when the temperature is raised to $30.0°\text{C}$.

is $580\text{N}$.

To determine

The fundamental frequency of the string.

Answer to Problem 19.73CP

The fundamental frequency of the string is $192\text{Hz}$.

Explanation of Solution

Given Info: The diameter of the string of the steel guitar is $1.00\text{mm}$, the string is stretched between the supports $80.0\text{cm}$ apart, the initial temperature is $0.0°\text{C}$, the density of the steel string is $7.86\times {10}^3\text{Kg}/{\text{m}}^{\text{3}}$, the fundamental frequency of the transverse oscillations of the string is $200\text{Hz}$, the temperature is raised to $30.0°\text{C}$

Write the expression for the fundamental frequency.

$f=\frac{1}{2L_2}\sqrt{\frac{T}{\mu }}$

Substitute $582\text{N}$ for $T$, $0.80026\text{m}$ for $L_2$, $6.17\times {10}^{-3}\text{Kg}/\text{m}$ for $\mu$ in the above expression.

$\begin{array}{c}f=\frac{1}{2\left(0.80026\text{m}\right)}\sqrt{\frac{\left(582\text{N}\right)}{\left(6.17\times {10}^{-3}\text{Kg}/\text{m}\right)}}\\ =192\text{Hz}\end{array}$

Conclusion:

Therefore, the fundamental frequency of the string is $192\text{Hz}$.