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Expert-verified Found in: Page 471 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # The following four waves are sent along strings with the same linear densities (x is in meters and t is in seconds). Rank the waves according to (a) their wave speed and (b) the tension in the strings along which they travel, greatest first:(1)${{\mathbit{Y}}}^{{\mathbf{1}}}{\mathbf{=}}{\mathbf{\left(}}{\mathbf{3}}{\mathbit{m}}{\mathbit{m}}{\mathbf{\right)}}{\mathbf{}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{-}}{\mathbf{3}}{\mathbit{t}}{\mathbf{\right)}}\mathbf{,}$ (3)${{\mathbit{y}}}^{{\mathbf{3}}}{\mathbf{=}}{\mathbf{\left(}}{\mathbf{1}}{\mathbit{m}}{\mathbit{m}}{\mathbf{\right)}}{\mathbf{}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbf{\left(}}{\mathbf{4}}{\mathbit{x}}{\mathbf{-}}{\mathbit{t}}{\mathbf{\right)}}$ ,(2) ${{\mathbit{y}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{\left(}}{\mathbf{6}}{\mathbit{m}}{\mathbit{m}}{\mathbf{\right)}}{\mathbf{}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbf{\left(}}{\mathbf{2}}{\mathbit{x}}{\mathbf{-}}{\mathbit{t}}{\mathbf{\right)}}$, (4) ${{\mathbit{y}}}^{{\mathbf{4}}}{\mathbf{=}}{\mathbf{\left(}}{\mathbf{2}}{\mathbit{m}}{\mathbit{m}}{\mathbf{\right)}}{\mathbf{}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{-}}{\mathbf{2}}{\mathbit{t}}{\mathbf{\right)}}$.

1. The waves can be ranked according to their wave speed as ${V}_{1}>{V}_{4}>{V}_{2}>{V}_{3}$ (greatest first)
2. The waves can be ranked according to their tension in the string along which they travel as ${T}_{1}>{T}_{4}>{T}_{2}>{T}_{3}$ (greatest first)
See the step by step solution

## Step 1: Given

The four waves along the strings with the same linear densities are,

${Y}_{1}=3mm\right)\mathrm{sin}\left(x-3t\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{y}_{2}=\left(6mm\right)\mathrm{sin}\left(2x-t\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{y}_{3}=\left(1mm\right)\mathrm{sin}\left(4x-t\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{y}_{4}=\left(2mm\right)\mathrm{sin}\left(x-2t\right)$

## Step 2: Determining the concept

Use the concept of the equation of transverse wave and speed of a travelling wave. The wave speed on a stretched string gives the relation between speed and tension in the string.

Formulae are as follow:

$y={y}_{m}\mathrm{sin}\left(kx-\omega t\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}v=\frac{\omega }{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}v=\sqrt{\frac{T}{\mu }}$

Where, v is wave speed, T is tension in string, 𝝁 is mass per unit length, 𝝎 is angular frequency, k is wave number, t is time.

## Step 3: Determining to rank the waves according to their wave speed

(a)

Rank the waves according to their wave speed :

The equation of transverse wave is,

$y={y}_{m}\mathrm{sin}\left(kx-\omega t\right)...........................................................................\left(1\right)$

The speed of the travelling wave is,

$v=\frac{w}{k}$

The equation (i) is,

${y}_{1}=\left(3mm\right)\mathrm{sin}\left(x-3t\right)$

Compare this equation with equation (1), then the speed of the travelling wave is,

${v}_{1}=\frac{3}{1}\phantom{\rule{0ex}{0ex}}{v}_{1}=3$

The equation (ii) is,

${y}_{2}=\left(6mm\right)\mathrm{sin}\left(2x-t\right)$

Compare this equation with equation (1), then the speed of the travelling wave is,

${V}_{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}{v}_{2}=0.5$

The equation (iii) is,

${y}_{3}=\left(1mm\right)\mathrm{sin}94x-t\right)$

Compare this equation with equation (1), then the speed of the travelling wave is,

${V}_{3}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}{V}_{3}=0.25$

The equation (iv) is,

${y}_{4}=\left(2mm\right)\mathrm{sin}\left(x-2t\right)$

Compare this equation with equation (1), then the speed of the travelling wave is

${V}_{4}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}{V}_{4}=2$

Hence, the rank of the waves according to the wave speed is ${V}_{1}>{V}_{4}>{V}_{2}>{V}_{3}$ (greatest first).

## Step 4: Determining to rank the waves according to the tension in the string along which they travel

(b)

Rank the waves according to tension:

The wave speed on a stretched string is,

$V=\sqrt{\frac{T}{\mu }}\phantom{\rule{0ex}{0ex}}V\propto \sqrt{T}$

The speed on the stretched string is directly proportional to the tension in the string with the same linear density.

The speed on the stretched string for equation (i) is,

${v}_{1}\propto \sqrt{{T}_{1}}$

The speed on the stretched string for equation (ii) is,

${V}_{2}\propto \sqrt{{T}_{2}}$

The speed on the stretched string for equation (iii) is,

${V}_{3}\propto \sqrt{{T}_{3}}$

The speed on the stretched string for equation (i) is,

${V}_{4}\propto \sqrt{{T}_{4}}$

Hence, the rank of the waves according to their tension is ${T}_{1}>{T}_{4}>{T}_{2}>{T}_{3}$ (greatest first).

Therefore, the wave speed by using its expression and rank their values can be found, by using the expression of the speed on the stretched string, and can also find the tension in each string and rank their values. ### Want to see more solutions like these? 