Chapter 18, Problem 18.76AP

A nylon siring has mass 5.50 g and length L = 86.0 cm. The lower end is tied to the floor, and the upper end is tied to a small set of wheels through a slot in a track on which the wheels move (Fig. P18.76). The wheels have a mass that is negligible compared with that of the siring, and they roll without friction on the track so that the upper end of the string is essentially free. Figure P18.76 At equilibrium, the string is vertical and motionless. When it is carrying a small-amplilude wave, you may assume the string is always under uniform tension 1.30 N. (a) Find the speed of transverse waves on the siring, (b) The string's vibration possibilities are a set of standing-wave states, each with a node at the fixed bottom end and an antinode at the free top end. Find the node-antinode distances for each of the three simplest states, (c) Find the frequency of each of these states.

To determine

The speed of transverse waves on the string.

Answer to Problem 18.76AP

The speed of transverse waves on the string is $14.26\text{m}/\mathrm{s}$.

Explanation of Solution

Given info: The length of the string is $86.0\text{cm}$, the mass of string is $5.50\text{g}$ and the tension in string is $1.30\text{N}$.

The linear mass density of the string is,

$\mu =\frac{m}{l}$

Here,

$m$ is the mass of string.

$l$ is the length of the string.

Substitute $5.50\text{g}$ for $m$ and $86.0\text{cm}$ for $l$ in above equation.

$\begin{array}{c}\mu =\frac{5.50\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)}{86.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ =6.395\times {10}^{-3}\text{kg}/\text{m}\end{array}$

The formula for speed of string is,

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

Here,

$T$ is the tension in string.

Substitute $1.30\text{N}$ for $T$ and $6.395\times {10}^{-3}\text{kg}/\text{m}$ for $\mu$ in above equation.

$\begin{array}{c}v=\sqrt{\frac{1.30\text{N}}{6.395\times {10}^{-3}\text{kg}/\text{m}}}\\ =14.26\text{m}/\mathrm{s}\end{array}$

Conclusion:

Therefore, the speed of transverse waves on the string is $14.26\text{m}/\mathrm{s}$.

To determine

The node-antinodes distances for each of the three simplest states.

Answer to Problem 18.76AP

The node-antinodes distances are $0.86\text{m}$, $0.287\text{m}$, and $0.172\text{m}$ for the three cases simultaneously.

Explanation of Solution

Given info: The length of the string is $86.0\text{cm}$, the mass of string is $5.50\text{g}$ and the tension in string is $1.30\text{N}$.

Consider the first condition in which a node is formed at the fixed bottom and an antinode is at top.

For the above condition, the length of the string is,

$l=\frac{\lambda }{4}$

Here,

$\lambda$ is the wavelength.

Rewrite the above equation.

$\lambda =4l$

Substitute $86.0\text{cm}$ for $l$ in above equation.

$\begin{array}{c}\lambda =4\times 86.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =3.44\text{m}\end{array}$

The node-antinodes distance is,

$d=\frac{\lambda }{4}$ . (1)

Substitute $3.44\text{m}$ for $\lambda$ in above equation.

$\begin{array}{c}{d}_1=\frac{3.44\text{m}}{4}\\ =0.86\text{m}\end{array}$

Consider the second condition in which two node is formed and an antinode is at top.

For the above condition, the length of the string is,

$l=\frac{3\lambda }{4}$

Rewrite the above equation.

$\lambda =\frac{4l}{3}$

Substitute $86.0\text{cm}$ for $l$ in above equation.

$\begin{array}{c}\lambda =\frac{4}{3}\times 86.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =1.1467\text{m}\end{array}$

Thus, the wavelength is $1.1467\text{m}$ when two node is formed and an antinode is at top.

Substitute $1.1467\text{m}$ for $\lambda$ in equation (1).

$\begin{array}{c}{d}_2=\frac{1.1467\text{m}}{4}\\ =0.287\text{m}\end{array}$

Consider the third condition in which three node is formed and an antinode is at top.

For the above condition, the length of the string is,

$l=\frac{5\lambda }{4}$

Rewrite the above equation.

$\lambda =\frac{4l}{3}$

Substitute $86.0\text{cm}$ for $l$ in above equation.

$\begin{array}{c}\lambda =\frac{4}{5}\times 86.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.688\text{m}\end{array}$

Thus, the wavelength is $0.688\text{m}$ when three node is formed and an antinode is at top.

Substitute $0.688\text{m}$ for $\lambda$ in equation (1).

$\begin{array}{c}{d}_3=\frac{0.688\text{m}}{4}\\ =0.172\text{m}\end{array}$

Conclusion:

Therefore, the node-antinodes distances are $0.86\text{m}$, $0.287\text{m}$, and $0.172\text{m}$ for the three cases simultaneously.

To determine

The frequency of each conditions in part (b).

Answer to Problem 18.76AP

The frequencies are $4.145\text{Hz}$, $12.436\text{Hz}$ and $20.7\text{Hz}$ simultaneously.

Explanation of Solution

Given info: The length of the string is $86.0\text{cm}$, the mass of string is $5.50\text{g}$ and the tension in string is $1.30\text{N}$.

The formula of frequency is,

$f=\frac{v}{\lambda }$ (2)

Here,

$v$ is the speed of transverse wave in string.

For fist condition:

Substitute $14.26\text{m}/\mathrm{s}$ for $v$ and $3.44\text{m}$ for $\lambda$ in equation (2).

$\begin{array}{c}f=\frac{14.26\text{m}/\mathrm{s}}{3.44\text{m}}\\ =4.145\text{Hz}\end{array}$

For second condition:

Substitute $14.26\text{m}/\mathrm{s}$ for $v$ and $1.1467\text{m}$ for $\lambda$ in equation (2).

$\begin{array}{c}f=\frac{14.26\text{m}/\mathrm{s}}{1.1467\text{m}}\\ =12.436\text{Hz}\end{array}$

For third condition:

Substitute $14.26\text{m}/\mathrm{s}$ for $v$ and $0.688\text{m}$ for $\lambda$ in equation (2).

$\begin{array}{c}f=\frac{14.26\text{m}/\mathrm{s}}{0.688\text{m}}\\ =20.7\text{Hz}\end{array}$

Conclusion:

Therefore, the frequencies are $4.145\text{Hz}$, $12.436\text{Hz}$ and $20.7\text{Hz}$ simultaneously.