Chapter 12, Problem 82P

(a)

To determine

The speed of the transverse wave on the string.

Answer to Problem 82P

The speed of the transverse wave on the string is $\underset{\_}{360\text{m/s}}$.

Explanation of Solution

Write the expression for the speed of transverse wave

$v_t=\sqrt{\frac{F}{\mu }}$ (I)

Here, $v_t$ is the speed of transverse wave, $\mu$ is the linear density

Write the expression for linear density

$\mu =\frac{mass}{L}$ (III)

From the expression of density of cylinder it is possible to obtain an expression for mass. $\pi r^{2}L$ is the volume of cylinder.

The density of cylinder

$\begin{array}{c}\rho =\frac{mass}{volume}\\ =\frac{mass}{\pi r^{2}L}\end{array}$ (IV)

Here, $r$ is the radius of the cylinder, $L$ is the length of the cylinder

Hence from equation (III) the equation for mass of cylinder

$mass=\rho \pi r^{2}L$ (V)

Substitute equation (IV) in (III)

$\begin{array}{c}\mu =\frac{\rho \pi r^{2}L}{L}\\ =\rho \pi r^2\end{array}$

Thus, the expression of speed of transverse wave is

$v_t=\sqrt{\frac{F}{\rho \pi r^2}}$ (VI)

Conclusion:

Substitute $800\text{kg}\cdot {\text{m/s}}^2$ for $F$, $7860{\text{kg/m}}^3$ for $\rho$, and $0.0005\text{m}$ for $r$ in equation (VI)

$\begin{array}{c}{v}_t=\sqrt{\frac{800\text{kg}\cdot {\text{m/s}}^2}{7860{\text{kg/m}}^3\times 3.14\times {\left(0.0005\text{m}\right)}^2}}\\ =360\text{m/s}\end{array}$

Therefore, the speed of the transverse wave on the string is $\underset{\_}{360\text{m/s}}$.

(b)

To determine

The fundamental frequency of transverse wave.

Answer to Problem 82P

The fundamental frequency of transverse wave is $\underset{\_}{f_t=150\text{Hz}}$.

Explanation of Solution

Write the expression for frequency of transverse wave

$f_t=\frac{v_t}{\lambda }$ (VII)

Here, $f_t$ is the frequency of transverse wave, $\lambda$ is the wavelength

In a simplest standing wave vibration, distance between the nodes gives the length of the wire, and wavelength is the double of distance between nodes.

Here $1.2\text{m}$ is the distance between nodes, therefore the wavelength is

$\begin{array}{c}\lambda =2\times 1.2\text{m}\\ \text{=2}\text{.4}\text{m}\end{array}$

Conclusion:

Substitute $360\text{m/s}$ for $v_t$ , and $2.4\text{m}$ for $\lambda$ in equation (VII)

$\begin{array}{c}{f}_t=\frac{360\text{m/s}}{2.4\text{m}}\\ =150\text{Hz}\end{array}$

Therefore, the fundamental frequency of transverse wave is $\underset{\_}{f_t=150\text{Hz}}$.

(c)

To determine

The speed of longitudinal wave.

Answer to Problem 82P

The speed of longitudinal wave is $\underset{\_}{5.04\text{km/s}}$.

Explanation of Solution

Write the expression for speed of longitudinal wave

$v_l=\sqrt{\frac{Y}{\rho }}$ (VIII)

Here, $Y$ is the young’s modulus, $\rho$ is the density

Conclusion:

Substitute $200\times {10}^9{\text{N/m}}^2$ for $Y$ , and $7860{\text{kg/m}}^2$ for $\rho$ in equation (VIII)

$\begin{array}{c}{v}_l=\sqrt{\frac{200\times {10}^9{\text{N/m}}^2}{7860{\text{kg/m}}^2}}\\ =5.04\text{km/s}\end{array}$

Therefore, speed of longitudinal wave is $\underset{\_}{5.04\text{km/s}}$.

(d)

To determine

The fundamental frequency of transverse wave.

Answer to Problem 82P

The

Explanation of Solution

Write the expression for frequency of longitudinal wave

$f_l=\frac{v_l}{\lambda }$ (IX)

Here, $f_l$ is the frequency of longitudinal wave, $\lambda$ is the wavelength

In a simplest standing wave vibration, distance between the nodes gives the length of the wire, and wavelength is the double of distance between nodes.

Here $1.2\text{m}$ is the distance between nodes, therefore the wavelength is

$\begin{array}{c}\lambda =2\times 1.2\text{m}\\ \text{=2}\text{.4}\text{m}\end{array}$

Conclusion:

Substitute $5044\text{m/s}$ for $v_t$ , and $2.4\text{m}$ for $\lambda$ in equation (VIII)

$\begin{array}{c}{f}_t=\frac{5044\text{m/s}}{2.4\text{m}}\\ =2.10\text{kHz}\end{array}$

Therefore, the fundamental frequency of longitudinal wave is $\underset{\_}{f_l=2.10\text{kHz}}$.