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Q9P

Expert-verifiedFound in: Page 472

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 **

**(a) What is ${{\mathit{y}}}_{{\mathbf{m}}}$ ****, **

**(b) What is k ****, **

**(c) What is ${\mathit{\omega}}$ ****, and **

**(d) What is the correct choice of sign in front of ${\mathit{\omega}}$****?**

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

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

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

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

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

**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, ${y}_{m}=\frac{1}{2}\times \left(peaktopeakdis\mathrm{tan}ce\right)$ (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)

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

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

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

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

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,$\lambda =40.0cmor40.0\times {10}^{-2}m$

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

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

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

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:

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

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

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

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

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

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