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Found in: Page 472

### Fundamentals Of Physics

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 ${{\mathbit{x}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{80}}{\mathbf{}}{\mathbit{m}}$ .What is the amplitude of the wave?

The amplitude of the wave is 0.2 m

See the step by step solution

## Step 1: The given data

1. The scale of the x axis = 0.80 m

## Step 2: Understanding the concept of wave equation

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, $y={y}_{m}\mathrm{cos}\left(kx-\omega t\right)$ (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

## Step 3: Calculation for the amplitude of the wave

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-

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

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.