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### Fundamentals Of Physics

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

# A sinusoidal wave moving along a string is shown twice in Figure 16-33, as crest A travels in the positive direction of an x axis by distance d = 6.0 cm in 4.0 ms . The tick marks along the axis are separated by 10 cm ; height H = 6.00mm If the wave equation is of the form, ${\mathbit{y}}\left(x,t\right){\mathbf{=}}{{\mathbit{y}}}_{{\mathbf{m}}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}\left(kx+\omega t\right)$(a) What is ${{\mathbit{y}}}_{{\mathbf{m}}}$ , (b) What is k , (c) What is ${\mathbit{\omega }}$ , and (d) What is the correct choice of sign in front of ${\mathbit{\omega }}$?

a) The maximum amplitude is $3.00×{10}^{-3}m$.

b) The wave vector is$16.0{m}^{-1}$ .

c) The angular frequency is $2.4×{10}^{2}rad/s$.

d) The correct choice of sign in front of ω is negative.

See the step by step solution

## Step 1: The given data

• The distance travelled by the crest,d = 6.0 cm or 0.06 m .
• The time required for this displacement,$t=4.0msor4.0×{10}^{-3}s$ .
• The scale on x -axis, the distance between the tick marks is10 cm .
• The peak-to-peak distance,$H=6.00mmor6.00×{10}^{-3}m$ .

## Step 2: Understanding the concept of wave equation

A sinusoidal wave traveling in positive x-direction is described by a standard equation. We use the equation and the information from the graph to calculate the required quantities.

Formula:

The maximum amplitude of the wave, ${y}_{m}=\frac{1}{2}×\left(peaktopeakdis\mathrm{tan}ce\right)$ (i)

The transverse speed of a wave, $v=\lambda f=\frac{\omega }{k}$ (ii)

The wavenumber of the wave, $k=\frac{2\mathrm{\pi }}{\lambda }$ (iii)

The velocity of a body in motion, $v=\frac{d}{t}$ (iv)

## Step 3: a) Calculation of the value of ym

It is given that the peak-to-peak distance is$H=6.00×{10}^{-3}m$.

The maximum displacement is given using equation (i) and the given values as follows:

${y}_{m}=\frac{1}{2}×6.00×{10}^{-3}\phantom{\rule{0ex}{0ex}}=3.00×{10}^{-3}m$

Hence, the value of maximum amplitude is $3.00×{10}^{-3}m$ .

## Step 4: b) Calculation of wavevector of the wave

From the graph, we can observe that the graph repeats itself after travelling a distance of 4 tick marks i.e. distance of (4 X 10 cm) = 40 cm

So we get,$\lambda =40.0cmor40.0×{10}^{-2}m$

Now using equation (iii) and given values, the wavevector can be given as:

$K=\frac{2×3.14}{40.0×{10}^{-2}}\phantom{\rule{0ex}{0ex}}\approx 16.0{m}^{-1}$

Hence, the value of wave vector is$16.0{m}^{-1}$ .

## Step 5: c) Calculation of angular frequency

Crest A moves distance d in time t in the positive direction of x axis. Thus, the wave velocity using equation (iv) is given as:

$v=\frac{6.0×{10}^{-2}}{4.0×{10}^{-3}}\phantom{\rule{0ex}{0ex}}=15m/s$

For a travelling wave, the wave velocity using equation (ii) is given as:

$\omega =vk\phantom{\rule{0ex}{0ex}}=15×16\phantom{\rule{0ex}{0ex}}=240\phantom{\rule{0ex}{0ex}}=2.4×{10}^{2}rad/s$

Hence, the value of angular velocity is$2.4×{10}^{2}rad/s$ .

## Step 6: d) Finding the sign of angular velocity

The sign of $\omega$ should be negative as the wave is moving along the positive direction of x axis.