A sinusoidal wave is sent along a string with a linear density of 2.0 g/m. As it travels, the kinetic energies of the mass elements along the string vary. Figure (a)gives the rate$\mathit{d}\mathit{K}\mathbf{/}\mathit{d}\mathit{t}$at which kinetic energy passes through the string elements at a particular instant, plotted as a function of distance x along the string. Figure (b)is similar except that it gives the rate at which kinetic energy passes through a particular mass element (at a particular location), plotted as a function of time t. For both figures, the scale on the vertical (rate) axis is set by Rs = 10 W. What is the amplitude of the wave?

1. Linear density,$\mu =2.0\mathrm{g}/\mathrm{m}\mathrm{or}0.002\mathrm{kg}/\mathrm{m}$
2. From figure, the kinetic energy rate,$\frac{dk}{dt}=10\mathrm{W}$

For a sinusoidal wave, by using the formula of average power, we can calculate the amplitude of the wave.

Formula:

The rate of kinetic energy, $\frac{dk}{dt}=\frac{1}{2}\mu \nu {\omega }^2{y}_m^2{\cos }^2\left(kx-\omega t\right)$(i)

The angular frequency of a body$\omega =2\pi f$ (ii)

The velocity of a body, $\nu =f\lambda$ (iii)

In equation (i), it can be seen that the rate of change of kinetic energy is the function of ${\cos }^2\left(kx-\omega t\right)$. So, using the relationship between the time period of $\cos$and ${\cos }^2$functions and using the time period given in the graph, we can write the frequency as:

$$\begin{gathered} f=\frac{1}{2\times {10}^{-3}\mathrm{s}} \\ =500\mathrm{Hz} \end{gathered}$$

Similarly, the velocity of the wave is given as:

$$\begin{gathered} v=\frac{0.1\mathrm{m}}{1\times {10}^{-3}\mathrm{s}} \\ =100\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (iii), the wavelength of a wave is given as:

$$\begin{gathered} \lambda =v/f \\ =\frac{100}{500} \\ =0.20\mathrm{m} \end{gathered}$$

From equation (i), we can writethat the rate of change of kinetic energy would bemaximum if ${\cos }^2\left(kx-\omega t\right)=1$

localid="1660983930043" $$\begin{gathered} \frac{dk}{dt}=\frac{1}{2}\mu \nu {\omega }^2{y}_m^2 \\ 10=\frac{1}{2}\mu \nu {\omega }^2{y}_m^2 \\ {y}_m^2=\frac{2\times 10}{2{\pi }^2\mu \lambda f^3}\left(\because fromgiven\right) \\ {y}_m^2=\frac{2\times 10}{\mu \nu {\omega }^2} \\ {y}_m=\sqrt{\frac{10}{2{\pi }^2\mu \lambda f^3}}\left(\because fromequation\left(ii\right)\right) \\ =\sqrt{\frac{10}{2\times 9.86\times 0.002\times 0.20\times {500}^3}} \\ =\sqrt{\frac{10}{986000}} \\ =0.0032\mathrm{m} \end{gathered}$$

Hence, the value of the amplitude is 0.0032 m