Chapter 12, Problem 62P

To determine

The weight should she hang from the wire if the temperature is $18.0°\text{C}$.

Answer to Problem 62P

The weight should she hang from the wire is $1,280\text{N}$.

Explanation of Solution

Write the expression for the fundamental frequency of the tube.

$f_1=\frac{v_{\text{air}}}{4L_{\text{tube}}}$ (I)

Here, $f_1$ is the fundamental frequency of the wire, $v_{\text{air}}$ is the speed of sound in air and $L_{\text{tube}}$ is the length of the tube.

Write the expression for the velocity of air with temperature.

$v_{\text{air}}=v_{\text{0}}\sqrt{\frac{T}{T_0}}$

Here, $v_0$ is the speed of sound at $0°\text{C}\text{or}\text{273}\text{.15}\text{K}$ and $T$ is the present temperature and $T_0$ is the $0°\text{C}\text{or}\text{273}\text{.15}\text{K}$.

Substitute $v_{\text{0}}\sqrt{\frac{T}{T_0}}$ for $v_{\text{air}}$ in equation (I) to get $f_1$.

$f_1=\frac{v_{\text{0}}}{4L_{\text{tube}}}\sqrt{\frac{T}{T_0}}$ (II)

Write the expression for appropriate wave speed of the wire.

$f_1=\frac{v_{\text{wire}}}{2L_{\text{wire}}}$ (III)

Here, $v_{\text{wire}}$ is the speed of sound in wire and $L_{\text{wire}}$ is the length of wire.

Rearrange above equation to get $v_{\text{wire}}$.

$v_{\text{wire}}=2L_{\text{wire}}{f}_1$

Write the expression for the linear mass density of wire.

$\mu =\frac{m_{\text{wire}}}{L_{\text{wire}}}$ (IV)

Here, $\mu$ is the linear mass density of the wire, $m_{\text{wire}}$ is the mass of the wire and $L_{\text{wire}}$ is the length of the wire.

Write the expression for the mass of the wire.

$m_{\text{wire}}={\rho }_{\text{wire}}{A}_{\text{wire}}{L}_{\text{wire}}$

Here, $m_{\text{wire}}$ is the mass of the wire, ${\rho }_{\text{wire}}$ is the density of the wire, $A_{\text{wire}}$ is the area of cross section of the wire.

Substitute ${\rho }_{\text{wire}}{A}_{\text{wire}}{L}_{\text{wire}}$ for $m_{\text{wire}}$ in equation (IV) to get $\mu$.

$\begin{array}{c}\mu =\frac{{\rho }_{\text{wire}}{A}_{\text{wire}}{L}_{\text{wire}}}{L_{\text{wire}}}\\ ={\rho }_{\text{wire}}{A}_{\text{wire}}\end{array}$

Write the expression for the weight to be hung from wire.

$v_{\text{wire}}=\sqrt{\frac{F}{\mu }}$

Substitute ${\rho }_{\text{wire}}{A}_{\text{wire}}$ for $\mu$ in above equation to get $v_{\text{wire}}$.

$v_{\text{wire}}=\sqrt{\frac{F}{{\rho }_{\text{wire}}{A}_{\text{wire}}}}$

Conclusion:

Rearrange above equation to get $F$.

$F={\rho }_{\text{wire}}{A}_{\text{wire}}{v}_{\text{wire}}^2$

Substitute $2L_{\text{wire}}{f}_1$ for $v_{\text{wire}}$ in above equation to get $v_{\text{wire}}$.

$F={\rho }_{\text{wire}}{A}_{\text{wire}}{\left(2L_{\text{wire}}{f}_1\right)}^2$

Substitute $\frac{v_{\text{0}}}{4L_{\text{tube}}}\sqrt{\frac{T}{T_0}}$ for $f_1$ and $\pi r^2$ for $A_{\text{wire}}$ in above equation to get $F$.

$F={\rho }_{\text{wire}}\pi r_{\text{wire}}^2{\left(\frac{L_{\text{wire}}}{L_{\text{tube}}}\right)}^2\frac{v_{\text{0}}^{2}T}{4T_0}$ (V)

Convert $18.0°\text{C}$ into Kelvin scale.

$\begin{array}{c}18.0°\text{C}=\left(\text{18}\text{.0+273}\text{.15}\right)\text{K}\\ =\text{291}\text{.15}\text{K}\end{array}$

Substitute $7860\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{wire}}$, $0.00200\text{m}$ for $r_{\text{wire}}$, $1.00\text{m}$ for $L_{\text{wire}}$,$1.\text{50}\text{m}$ for $L_{\text{tube}}$, $291.15\text{K}$ for $T$ , $273.15\text{K}$ for $T_0$ and $331.0\text{m}/\text{s}$ for $v_0$ in equation (V).

$\begin{array}{c}F=\left(7860\text{kg}/{\text{m}}^3\right)\pi {\left(0.00200\text{m}\right)}^2{\left(\frac{1.00\text{m}}{1.\text{50}\text{m}}\right)}^2\frac{{\left(331.0\text{m}/\text{s}\right)}^2\left(291.15\text{K}\right)}{4\left(273.15\text{K}\right)}\\ =1280\text{N}\end{array}$

Therefore, the weight should she hang from the wire is $1,280\text{N}$.