Chapter 15, Problem 103P

(a)

To determine

To calculate:The wave speed.

Answer to Problem 103P

The wave speed $\left(\text{v}\right)$ is $\text{10}\text{.0}\text{m/s}$ .

Explanation of Solution

Given:

Mass per unit length of the rope = $\text{0}\text{.100}\text{kg/m}$ .

Tension = $\text{10}\text{.0}\text{N}$ .

Frequency = $\text{5}\text{.00}$ cycles per second.

Amplitude = $\text{40}\text{.0}\text{mm}$ .

Formula used:

The wave speed can be calculated as:

  $\text{v=}\sqrt{\frac{{\text{F}}_{\text{T}}}{\text{μ}}}$

Where,

  $\text{v=}$ Velocity of the wave.

  ${\text{F}}_{\text{T}}=$ tension in the wave.

  $\text{μ=}$ Mass per unit length of the rope.

Calculation:

Let, the $\text{Δm}$ represents the mass of the segment of the length $\text{Δx=1}\text{.00}\text{mm}$ .

From the given data, wave speed can be found out as:

  $\text{v=}\sqrt{\frac{{\text{F}}_{\text{T}}}{\text{μ}}}$

Where,

  $\text{v=}$ Velocity of the wave.

  ${\text{F}}_{\text{T}}=$ tension in the wave.

  $\text{μ=}$ Mass per unit length of the rope.

Now, substituting the all values of the linear density and tension in the equation:

  $\begin{array}{l}\text{v=}\sqrt{\frac{{\text{F}}_{\text{T}}}{\text{μ}}}\\ \text{v=}\sqrt{\frac{\text{10}\text{.0}\text{N}}{\text{0}\text{.100}\text{kg/m}}}\\ \text{v=10}\text{.0}\text{m/s}\end{array}$

Conclusion:

Thus, the wave speed $\left(\text{v}\right)$ is $\text{10}\text{.0}\text{m/s}$ .

(b)

To determine

To calculate: The wavelength.

Answer to Problem 103P

The wavelength $\left(\text{λ}\right)$ is $\text{2}\text{.00}\text{m}$ .

Explanation of Solution

Given:

Mass per unit length of the rope = $\text{0}\text{.100}\text{kg/m}$ .

Tension = $\text{10}\text{.0}\text{N}$ .

Frequency = $\text{5}\text{.00}$ cycles per second.

Amplitude = $\text{40}\text{.0}\text{mm}$ .

Formula used:

The wave speed can be calculated as:

  $\text{v= f λ}$

Where,

Sound’s speed: $\text{v}$ .

Frequency of wave: $\text{f }$ .

The wavelength: $\text{λ}$ .

Calculation:

To evaluate the wavelength, expression used is:

  $\text{v= f λ}$

Where,

Sound’s speed: $\text{v}$ .

Frequency of wave: $\text{f }$ .

The wavelength: $\text{λ}$ .

After substituting the values of the

Now, substituting the all values of the linear density and tension in the equation:

  $\begin{array}{l}\text{λ=}\frac{\text{v}}{\text{f }}\\ \text{λ=}\frac{\text{10}\text{.0}\text{m/s}}{\text{5}\text{.00}{\text{s}}^{\text{-1}}}\\ \text{λ=2}\text{.00}\text{m}\end{array}$

Conclusion:

Thus, the wavelength $\left(\text{λ}\right)$ is $\text{2}\text{.00}\text{m}$ .

(c)

To determine

To calculate:The maximum transverse linear momentum.

Answer to Problem 103P

The maximum transverse linear momentum $\left({\text{p}}_{\text{max}}\right)$ is $\text{1}{\text{.26×10}}^{\text{-4}}\text{kg}\text{.m/s}$ .

Explanation of Solution

Given:

Mass per unit length of the rope = $\text{0}\text{.100}\text{kg/m}$ .

Tension = $\text{10}\text{.0}\text{N}$ .

Frequency = $\text{5}\text{.00}$ cycles per second.

Amplitude = $\text{40}\text{.0}\text{mm}$ .

Formula used:

To calculate the maximum transverse linear momentum the expression used is:

  ${\text{p}}_{\text{max}}{\text{=Δmv}}_{\text{max}}$

Calculation:

Initially, relate the max transverse linear momentum of the $\text{1}\text{.00}\text{mm}$ segment to the max transverse speed of the wave.

  $\begin{array}{l}{\text{p}}_{\text{max}}{\text{=Δmv}}_{\text{max}}\\ {\text{p}}_{\text{max}}\text{=μΔxAω}\\ {\text{p}}_{\text{max}}\text{=2πfμΔxA}\end{array}$

Substitute the numerical values to evaluate ${\text{p}}_{\text{max}}$:

  $\begin{array}{l}{\text{p}}_{\text{max}}\text{=2πfμΔxA}\\ {\text{p}}_{\text{max}}\text{=2π}\left(\text{5}\text{.00}{\text{s}}^{\text{-1}}\right)\left(\text{0}\text{.100}\text{kg/m}\right)\text{×}\left(\text{1}{\text{.00×10}}^{\text{-3}}\text{m}\right)\left(\text{0}\text{.0400}\text{m}\right)\\ {\text{p}}_{\text{max}}\text{=1}{\text{.257×10}}^{\text{-4}}\text{kg}\text{.m/s}\\ {\text{p}}_{\text{max}}\text{=1}{\text{.26×10}}^{\text{-4}}\text{kg}\text{.m/s}\end{array}$

Conclusion:

Thus, the maximum transverse linear momentum $\left({\text{p}}_{\text{max}}\right)$ is $\text{1}{\text{.26×10}}^{\text{-4}}\text{kg}\text{.m/s}$ .

(d)

To determine

To calculate: The maximum net force.

Answer to Problem 103P

The maximum net force $\left({\text{F}}_{\text{max}}\right)$ is $\text{3}\text{.95}\text{mN}$ .

Explanation of Solution

Given:

Mass per unit length of the rope = $\text{0}\text{.100}\text{kg/m}$ .

Tension = $\text{10}\text{.0}\text{N}$ .

Frequency = $\text{5}\text{.00}$ cycles per second.

Amplitude = $\text{40}\text{.0}\text{mm}$ .

Formula used:

To calculate the maximum net forcethe expression used is:

  ${\text{F}}_{\text{max}}{\text{=Δma}}_{\text{max}}$

Calculation:

The max net force acting on the segment is the product of the mass of th segment and its max acceleration:

  $\begin{array}{l}{\text{F}}_{\text{max}}{\text{=Δma}}_{\text{max}}\\ {\text{F}}_{\text{max}}{\text{=μΔxAω}}^{\text{2}}\\ {\text{F}}_{\text{max}}{\text{=ωp}}_{\text{max}}\\ {\text{F}}_{\text{max}}{\text{=2πfp}}_{\text{max}}\end{array}$

Substitute the numerical values to evaluate ${\text{F}}_{\text{max}}$ :

  $\begin{array}{l}{\text{F}}_{\text{max}}{\text{=2πfp}}_{\text{max}}\\ {\text{F}}_{\text{max}}\text{=2π}\left(\text{5}\text{.00}{\text{s}}^{\text{-1}}\right)\left(\text{1}{\text{.257×10}}^{\text{-4}}\text{kg}\text{.m/s}\right)\\ {\text{F}}_{\text{max}}\text{=3}\text{.95}\text{mN}\end{array}$

Conclusion:

Thus, the maximum net force $\left({\text{F}}_{\text{max}}\right)$ is $\text{3}\text{.95}\text{mN}$ .