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Found in: Page 473

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The equation of a transverse wave on a string is ${\mathbit{y}}{\mathbf{=}}\left(2.0mm\right){\mathbit{s}}{\mathbit{i}}{\mathbit{n}}\left[\left(20{m}^{-1}\right)x-\left(600{s}^{-1}\right)t\right]$ . The tension in the string is 15 N . (a)What is the wave speed? (b)Find the linear density of this string in grams per meter.

a) The speed of the wave is $30m/s$

b) The linear density of this string is $1.7×{10}^{-2}kg/m$

See the step by step solution

## Step 1: The given data

• Equation of a transverse wave on a string is- $y=\left(2.0mm\right)\mathrm{sin}\left[\left(20{m}^{-1}\right)x-\left(600{s}^{-1}\right)t\right]$
• Tension in the string, T = 15 N
• Angular frequency of the wave, $\omega =600{s}^{-1}$
• Wavelength of the wave, $\lambda =0.05m$

## Step 2: Understanding the concept of the wave equation

The product of the frequency and wavelength of the wave, gives us the wave speed. The linear density is the term used in place of the mass per unit length of the string.

Formula:

The speed of the wave, $v=n×\lambda$ (i)

The formula for the linear density of distribution, $d=\frac{T}{{v}^{2}}$ (ii)

## Step 3: a) Calculation for the wave speed

Using the equation (i) and the required given values, the wave speed is given as:

$v=600{s}^{-1}×0.05m\phantom{\rule{0ex}{0ex}}=30m/s$

Hence, the value of the wave speed is

## Step 4: b) Calculation of the linear density

Using equation (i) and the value of wave speed, we get the linear density as:

$d=\frac{15N}{{\left(30m/s\right)}^{2}}\phantom{\rule{0ex}{0ex}}=1.6667×{10}^{-2}kg/m\phantom{\rule{0ex}{0ex}}\approx 1.7×{10}^{-2}kg/m$

Hence, the value of the linear density of the string is $1.7×{10}^{-2}kg/m$