Chapter 19, Problem 73CP

(a)

To determine

The mas per unit length of the string.

Answer to Problem 73CP

The mas per unit length of the string is $\underset{\_}{6.17\times {10}^{-3}\text{kg/m}}$.

Explanation of Solution

Write the equation for the mass per unit length.

  $\mu =\frac{1}{4}\rho \left(\pi d^2\right)$                                                                                                (I)

Here, $\mu$ is the mass per unit length, $\rho$ is the density and $d$ is the diameter.

Conclusion:

Substitute $7.86\times {10}^3{\text{kg/m}}^{\text{3}}$ for $\rho$, $1.00\text{mm}$ for $d$ in Equation (I) to find $\mu$.

  $\begin{array}{c}\mu =\frac{1}{4}\left(7.86\times {10}^3{\text{kg/m}}^{\text{3}}\right)\left(\pi {\left\{\left(1.00\text{mm}\right)\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)\right\}}^2\right)\\ =\frac{1}{4}\left(7.86\times {10}^3{\text{kg/m}}^{\text{3}}\right)\left(\pi {\left(1.00\times {10}^{-3}\text{m}\right)}^2\right)\\ =6.17\times {10}^{-3}\text{kg/m}\end{array}$

Thus, the mas per unit length of the string is $\underset{\_}{6.17\times {10}^{-3}\text{kg/m}}$.

(b)

To determine

The tension in the string.

Answer to Problem 73CP

The tension in the string is $\underset{\_}{632\text{N}}$.

Explanation of Solution

Write the equation for the fundamental frequency.

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

Here, $f_1$ is the fundamental frequency, $v$ is the speed and $L$ is the length.

Write the equation for the speed.

  $v=\sqrt{\frac{T}{\mu }}$                                                                                                 (III)

Here, $T$ is the tension.

Rewrite the expression for the tension by using (II) and (III).

  $T=\mu {\left(2L{f}_1\right)}^2$                                                                                       (IV)

Conclusion:

Substitute $6.17\times {10}^{-3}\text{kg/m}$ for $\mu$, $80.0\text{cm}$ for $L$, $200\text{Hz}$ for $f_1$ in Equation (IV) to find $T$.

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

Thus, the tension in the string is $\underset{\_}{632\text{N}}$.

(c)

To determine

The tension in the string at $30.0\text{°C}$.

Answer to Problem 73CP

The tension in the string at $30.0\text{°C}$ is $\underset{\_}{580\text{N}}$.

Explanation of Solution

Write the equation for actual length.

  $L_{\text{actual}}=L_{0°C}\left(1+\frac{F}{AY}\right)$                                                                                     (V)

Here, $L_{\text{actual}}$ is the actual length, $L_{0\text{°C}}$ is the length at $0\text{°C}$, $F$ is the tension, $A$ is the cross-sectional area and $Y$ is the Young’s modulus.

Write the equation for the area of cross-section.

  $A=\left(\frac{\pi d^2}{4}\right)$                                                                                                (VI)

Here, $d$ is the diameter.

Rewrite the expression for the length at $0\text{°C}$ by using (V) and (VI).

  $L_{0°C}=\frac{L_{\text{actual}}}{\left[1+\frac{F}{\left(\pi d^2/4\right)Y}\right]}$                                                                          (VII)

Write the expression for the unstretched length at $30\text{°C}$.

  $L_{30\text{°C}}=L{\text{0°C}}_{}\left[1+\alpha \left(T_2-T_1\right)\right]$                                                                         (VIII)

Here, $L_{30\text{°C}}$ is the length at $30\text{°C}$, $\alpha$ is the expansion coefficient, $T_2$ is the final temperature and $T_1$ is the initial temperature.

Write the expression for the tension at $30\text{°C}$ by using equation (V) and (VIII).

  $F^{\prime }=\left(\frac{\pi d^2}{4}\right)Y\left[\frac{L_{actual}}{L{\text{0°C}}_{}\left[1+\alpha \left(T_2-T_1\right)\right]}-1\right]$                                                (IX)

Here, $F^{\prime }$ is the tension at $30\text{°C}$.

Conclusion:

Substitute $1.00\text{mm}$ for $d$, $80.0\text{cm}$ for $L_{\text{actual}}$, $632\text{N}$ for $F$ and $20.0\times {10}^{10}{\text{N/m}}^{\text{2}}$ for $Y$ in Equation (VII) to find $L_{0\text{°C}}$.

  $\begin{array}{c}{L}_{0°C}=\frac{\left[\left(80.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left[1+\frac{\left(632\text{N}\right)}{\left(\frac{\pi }{4}{\left\{\left(1.00\text{mm}\right)\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)\right\}}^2\right)\left(20.0\times {10}^{10}{\text{N/m}}^{\text{2}}\right)}\right]}\\ =0.7968\text{m}\end{array}$

Substitute $1.00\text{mm}$ for $d$, $80.0\text{cm}$ for $L_{\text{actual}}$, $0.7968\text{m}$ for $L_{0\text{°C}}$, $11.0\times {10}^{-6}\text{/°C}$ for $\alpha$, $30\text{°C}$ for $T_2$, $0\text{°C}$ for $T_1$ and $20.0\times {10}^{10}{\text{N/m}}^{\text{2}}$ for $Y$ in Equation (IX) to find $F\text{'}$.

  $\begin{array}{c}F\text{'}=\left[\left(\frac{\pi }{4}{\left\{\left(1.00\text{mm}\right)\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)\right\}}^2\right)\left(20.0\times {10}^{10}{\text{N/m}}^{\text{2}}\right)\right]\\ \left[\frac{\left[\left(80.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left(0.7968\text{m}\right)\left[1+\left(11.0\times {10}^{-6}\text{/°C}\right)\left(30\text{°C}-0\text{°C}\right)\right]}-1\right]\\ =580\text{N}\end{array}$

Thus, the tension in the string at $30.0\text{°C}$ is $\underset{\_}{580\text{N}}$.

(d)

To determine

The fundamental frequency.

Answer to Problem 73CP

The fundamental frequency is $\underset{\_}{192\text{Hz}}$.

Explanation of Solution

The frequency is directly proportional to the square root of tension.

Write the equation for new fundamental frequency.

  ${f^{\prime }}_1=f_1\sqrt{\frac{F^{\prime }}{F}}$                                                                                    (X)

Here, ${f^{\prime }}_1$ is the new fundamental frequency.

Conclusion:

Substitute $580\text{N}$ for $F^{\prime }$, $200\text{Hz}$ for $f_1$, $632\text{N}$ for $F$ in Equation (X) to find ${f^{\prime }}_1$.

  $\begin{array}{c}{f^{\prime }}_1=\left(200\text{Hz}\right)\sqrt{\frac{580\text{N}}{632\text{N}}}\\ =192\text{Hz}\end{array}$

Thus, the fundamental frequency is $\underset{\_}{192\text{Hz}}$.