Chapter 15, Problem 41P

(a)

To determine

To calculate:

Calculate the wave speed and its direction.

Answer to Problem 41P

The wave speed is $\text{5}\text{.00}\text{m/s}$ and its waves travelled in $\text{-x}$ direction.

Explanation of Solution

Given:

Wave function: $\text{y}\left(\text{x,t}\right)\text{=}\left(\text{1}\text{.00}\text{mm}\right)\text{sin}\left(\text{62}\text{.8}{\text{m}}^{\text{-1}}\text{x+314}{\text{s}}^{\text{-1}}\text{t}\right)$ .

Formula used:

The speed of the wave is determined by:

  $\text{v=}\frac{\text{ω}}{\text{k}}$

Where,

  $\text{v=}$ Speed of the wave.

  $\text{ω=}$ Angular frequency of the wave

  $\text{k=}$ Spring constant.

Calculation:

The wave function $\text{y}\left(\text{x,t}\right)\text{=A sin }\left(\text{kx-ωt}\right)$ describes that the wave is travelling in the $\text{+x}$ direction.

For wave travelling in $\text{-x}$ direction, wave function is $\text{y}\left(\text{x,t}\right)\text{=A sin }\left(\text{kx+ωt}\right)$ .By examination of the wave function, the $\text{A,K,ω}$ can be found out.

The wavelength, frequency and the period of the wave is determined from the $\text{K,ω}$ .

As the given wave function is,

  $\text{y}\left(\text{x,t}\right)\text{=}\left(\text{1}\text{.00}\text{mm}\right)\text{sin}\left(\text{62}\text{.8}{\text{m}}^{\text{-1}}\text{x+314}{\text{s}}^{\text{-1}}\text{t}\right)$

And because of the sign in between the $\text{kx}$ and $\text{ωt}$ terms is positive, so the wave travelling in the $\text{-x}$ direction.

The speed of the wave is,

  $\begin{array}{l}\text{v=}\frac{\text{ω}}{\text{k}}\\ \text{v=}\frac{\text{314}{\text{s}}^{\text{-1}}}{\text{62}\text{.8}{\text{m}}^{\text{-1}}}\\ \text{v=5}\text{.00}\text{m/s}\end{array}$

Thus, $\text{v=5}\text{.00}\text{m/s}$

Conclusion:

Thus, the wave speed is $\text{5}\text{.00}\text{m/s}$ and its waves travelled in $\text{-x}$ direction.

(b)

To determine

To calculate:

Calculate the wavelength, frequency and period of the wave.

Answer to Problem 41P

The wavelength $\left(\text{λ}\right)$ , frequency $\left(\text{f}\right)$ and period $\left(\text{T}\right)$ of the wave is $\text{10}\text{.0}\text{cm}$ , $\text{50}\text{.0}\text{Hz}$ , and $\text{0}\text{.0200}\text{s}$ respectively.

Explanation of Solution

Given:

Wave function: $\text{y}\left(\text{x,t}\right)\text{=}\left(\text{1}\text{.00}\text{mm}\right)\text{sin}\left(\text{62}\text{.8}{\text{m}}^{\text{-1}}\text{x+314}{\text{s}}^{\text{-1}}\text{t}\right)$ .

Formula used:

The wavelength is determined by:

  $\text{λ=}\frac{\text{2π}}{\text{k}}$

Where,

  $\text{λ=}$ Wavelength of the speed

  $\text{k=}$ Spring constant

The period is determined by:

  $\text{T=}\frac{\text{1}}{\text{f}}$

Where,

  $\text{T=}$ Time period of the wave.

  $\text{f=}$ Frequency

Calculation:

The wave function $\text{y}\left(\text{x,t}\right)\text{=A sin }\left(\text{kx-ωt}\right)$ describe the wave is travelling in the $\text{+x}$ direction.

For wave travelling in $\text{-x}$ direction, wave function is $\text{y}\left(\text{x,t}\right)\text{=A sin }\left(\text{kx+ωt}\right)$ .By examination of the wave function, the $\text{A,K,ω}$ can be found out.

The wavelength, frequency and the period of the wave is determined from the $\text{K,ω}$ .

The coefficient of the $\text{x}\text{is}\text{k}$ ,

Where,

  $\begin{array}{l}\text{k=}\frac{\text{2π}}{\text{λ}}\\ \text{λ=}\frac{\text{2π}}{\text{k}}\end{array}$

Now, substitute the numerical values to evaluate $\text{λ}$ ,

  $\begin{array}{l}\text{λ=}\frac{\text{2π}}{\text{k}}\\ \text{λ=}\frac{\text{2π}}{\text{62}\text{.8}{\text{m}}^{\text{-1}}}\\ \text{λ=10}\text{.0}\text{cm}\end{array}$

The coefficient of the $\text{t}$ and $\text{ω}$ is,

  $\begin{array}{l}\text{f=}\frac{\text{ω}}{\text{2π}}\\ \text{f=}\frac{\text{314}{\text{s}}^{\text{-1}}}{\text{2π}}\\ \text{f=50}\text{.0}\text{Hz}\end{array}$

The period of the wave motion is the reciprocal of its frequency,

  $\begin{array}{l}\text{T=}\frac{\text{1}}{\text{f}}\\ \text{T=}\frac{\text{1}}{\text{50}\text{.0}{\text{s}}^{\text{-1}}}\\ \text{T=0}\text{.0200}\text{s}\end{array}$

Conclusion:

Thus, the wavelength $\left(\text{λ}\right)$ , frequency $\left(\text{f}\right)$ and period $\left(\text{T}\right)$ of the wave is $\text{10}\text{.0}\text{cm}$ , $\text{50}\text{.0}\text{Hz}$ , and $\text{0}\text{.0200}\text{s}$ respectively.

(c)

To determine

To calculate:

Calculate the maximum speed of any point on the string.

Answer to Problem 41P

The maximum speed $\left({\text{v}}_{\text{max}}\right)$ of any point on the string is $\text{0}\text{.314}\text{m/s}$ .

Explanation of Solution

Given:

Wave function: $\text{y}\left(\text{x,t}\right)\text{=}\left(\text{1}\text{.00}\text{mm}\right)\text{sin}\left(\text{62}\text{.8}{\text{m}}^{\text{-1}}\text{x+314}{\text{s}}^{\text{-1}}\text{t}\right)$ .

Formula used:

The max velocity is calculated as:

  ${\text{v}}_{\text{max}}\text{=Aω}$

where,

  ${\text{v}}_{\text{max}}\text{=}$ Max velocity

  $\text{A=}$ Amplitude

  $\text{ω=}$ Angular frequency

Calculation:

To express and evaluate the maximum speed of any string segment is,

  $\begin{array}{l}{\text{v}}_{\text{max}}\text{=Aω}\\ {\text{v}}_{\text{max}}\text{=}\left(\text{1}\text{.00}\text{mm}\right)\left(\text{314}\text{rad/s}\right)\\ {\text{v}}_{\text{max}}\text{=0}\text{.314}\text{m/s}\end{array}$

Conclusion:

Thus, the maximum speed $\left({\text{v}}_{\text{max}}\right)$ of any point on the string is $\text{0}\text{.314}\text{m/s}$ .