Chapter 18, Problem 76AP

(a)

To determine

The speed of the transverse waves.

Answer to Problem 76AP

The speed of the transverse waves is $14.3\text{m/s}$.

Explanation of Solution

Write the expression for mass per unit length wire.

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

Here, $\mu$ is the linear density, $m$ is the mass of the nylon string, and $L$ is the length of the string.

Write the expression for tension in the string.

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

Here, $T$ is the tension and $\mu$ is the linear density.

Conclusion:

Substitute $5.50\text{g}$ for $m$ and $86.0\text{cm}$ for $L$ in equation (I) to find $\mu$.

  $\begin{array}{c}\mu =\frac{5.50\text{g}\left(\frac{0.001\text{kg}}{1\text{g}}\right)}{86.0\text{cm}\left(\frac{0.01\text{m}}{1\text{cm}}\right)}\\ =\frac{0.0055\text{kg}}{0.86\text{m}}\\ =0.0064\text{kg/m}\end{array}$

Substitute $0.005\text{kg/m}$ for $\mu$ and $1.30\text{N}$ for $T$ in equation (II) to find $v$.

  $\begin{array}{c}v=\sqrt{\frac{1.30\text{N}}{0.0064\text{kg/m}}}\\ =14.3\text{m/s}\end{array}$

Therefore, the speed of the transverse waves is $14.3\text{m/s}$.

(b)

To determine

Find the nodes and antinodes distance for three states.

Answer to Problem 76AP

The simplest pattern for one node and antinode distance is $86.0\text{cm}$, three node antinodes distance is $28.7\text{cm}$, and five node antinode pair is $17.2\text{cm}$.

Explanation of Solution

Since the distance between a node and antinode is one quarter of a wavelength. The string had contained only odd number of node antinode pairs.

Write the expression for simplest pattern AN one node antinode pair.

  $L=\frac{{\lambda }_1}{4}$                                                                                                          (III)

Here, $L$ is the length of one node antinode pair and ${\lambda }_1$ is the wavelength of the one nodal points.

Write the expression for simplest pattern ANAN three node antinode pairs.

  $L=3\frac{{\lambda }_3}{4}$                                                                                                       (IV)

Here, $L$ is the length of three node antinode pairs and ${\lambda }_3$ is the wavelength of three nodal points.

Write the expression for simplest pattern ANANAN five node antinode pairs.

  $L=5\frac{{\lambda }_5}{4}$                                                                                                          (V)

Here, $L$ is the length of five node antinode pairs and ${\lambda }_5$ is the wavelength of five nodal points.

Conclusion:

Substitute $86.0\text{cm}$  for ${\lambda }_1/4$ in equation (III) to find $L$.

  $L=86.0\text{cm}$

Substitute $86.0\text{cm}$  for ${\lambda }_3/4$ in equation (IV) to find $L$.

  $\begin{array}{c}\frac{L}{3}=\frac{{\lambda }_3}{4}\\ =28.7\text{cm}\end{array}$

Substitute $86.0\text{cm}$  for ${\lambda }_5/4$ in equation (V) to find $L$.

  $\begin{array}{c}\frac{L}{5}=\frac{{\lambda }_5}{4}\\ =17.2\text{cm}\end{array}$

Therefore, the simplest pattern for one node and antinode distance is $86.0\text{cm}$, three node antinodes distance is $28.7\text{cm}$, and five node antinode pair is $17.2\text{cm}$.

(c)

To determine

Find the frequency of nodes and antinodes for three states.

Answer to Problem 76AP

The frequency of the simplest pattern for one node and antinode distance is $4.14\text{Hz}$, three node antinodes distance is $12.4\text{Hz}$, and five node antinode pair is $20.7\text{Hz}$.

Explanation of Solution

Since the distance between a node and antinode is one quarter of a wavelength. The string had contained only odd number of node antinode pairs.

Write the expression for frequency of the simplest pattern AN one node antinode pair.

  $f_1=\frac{v}{4\left(\frac{{\lambda }_1}{4}\right)}$                                                                                                       (VI)

Here, $f_1$ is the frequency of the simplest pattern AN.

Write the expression for frequency of the simplest pattern ANAN three node antinode pair.

  $f_3=\frac{v}{4\left(\frac{{\lambda }_3}{4}\right)}$                                                                                                        (VII)

Here, $f_3$ is the frequency of the simplest pattern ANAN.

Write the expression for frequency of the simplest pattern ANANAN five node antinode pair.

  $f_5=\frac{v}{4\left(\frac{{\lambda }_5}{4}\right)}$                                                                                                       (VIII)

Here, $f_5$ is the frequency of the simplest pattern ANANAN.

Conclusion:

Substitute $86.0\text{cm}$ or ${\lambda }_1/4$ and $14.3\text{m/s}$ for $v$ in equation (VI) to find $f_1$.

  $\begin{array}{c}{f}_1=\frac{14.3\text{m/s}}{4\left(86.0\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)}\\ =\frac{14.3\text{m/s}}{4\left(0.860\text{m}\right)}\\ =4.14\text{Hz}\end{array}$

Substitute $28.7\text{cm}$ or ${\lambda }_3/4$ and $14.3\text{m/s}$ for $v$ in equation (VII) to find $f_3$.

  $\begin{array}{c}{f}_3=\frac{14.3\text{m/s}}{4\left(28.7\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)}\\ =\frac{14.3\text{m/s}}{4\left(0.287\text{m}\right)}\\ =12.4\text{Hz}\end{array}$

Substitute $17.2\text{cm}$ or ${\lambda }_5/4$ and $14.3\text{m/s}$ for $v$ in equation (VIII) to find $f_5$.

  $\begin{array}{c}{f}_5=\frac{14.3\text{m/s}}{4\left(17.2\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)}\\ =\frac{14.3\text{m/s}}{4\left(0.172\text{m}\right)}\\ =20.7\text{Hz}\end{array}$

Therefore, the frequency of the simplest pattern for one node and antinode distance is $4.14\text{Hz}$, three node antinodes distance is $12.4\text{Hz}$, and five node antinode pair is $20.7\text{Hz}$.