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Q6P

Expert-verifiedFound in: Page 472

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A sinusoidal wave travels along a string under tension. Figure 16-31 gives the slopes along the string at time t =0** **.The scale of the x** **axis is set by ${{\mathit{x}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{80}}{\mathbf{}}{\mathit{m}}$** **.What is the amplitude of the wave?**

The amplitude of the wave is 0.2 m

- The scale of the x axis = 0.80 m

**The sinusoidal wave exhibits different displacements at different positions .Thus, the slope at different points varies with ****the**** position. We use this concept along with the equation of the travelling wave to calculate amplitude.**

Formula:

The expression of wave equation, $y={y}_{m}\mathrm{cos}(kx-\omega t)$ (i)

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

Here, $\omega $ is the angular velocity of the wave and ${y}_{m}$ is the amplitude of the oscillation

The scale of x axis is given as 0.80 m. This distance is equivalent to two wavelengths.

Hence, the wavelength is given as:

$\begin{array}{l}2\lambda =0.80m\\ \lambda =0.40\mathrm{m}\end{array}$

For x=0 and t=0 , equation (i) becomes-

$\begin{array}{ll}y={y}_{m}& \mathrm{cos}(k\left(0\right)-\omega \left(0)\right)\\ y={y}_{m}& \mathrm{cos}\left(0\right)\\ y={y}_{m}& \phantom{\rule{0ex}{0ex}}\end{array}$

From the graph, the value of y at x =0 is 0.2 . So, the above equation becomes-

${y}_{m}=0.2$

Thus, the amplitude of the curve given in the graph is 0.2.

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