Question

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?

Short answer

The amplitude of the wave is 0.2 m

Step-by-step solution

The given data

1. The scale of the x axis = 0.80 m

Understanding the concept of wave equation

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

Formula:

The expression of wave equation, $y=y_m\cos \left(kx-\omega t\right)$ (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

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-

$\begin{array}{ll}y=y_m& \cos (k\left(0\right)-\omega \left(0\right))\\ y=y_m& \cos \left(0\right)\\ y=y_m& \end{array}$

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.