91Ó°ÊÓ

1. Linear density,$\mathrm{μ}=2.0\mathrm{g}/\mathrm{m}\mathrm{or}0.002\mathrm{kg}/\mathrm{m}$
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, $\frac{dk}{dt}=\frac{1}{2}\mathrm{μ}\mathrm{ν}{\mathrm{Ó\neg }}^2{y}_m^2{\cos }^2\left(kx-\mathrm{Ó\neg }t\right)$(i)

The angular frequency of a body$\mathrm{Ó\neg }=2\mathrm{Ï€}f$ (ii)

The velocity of a body, $\mathrm{ν}=f\mathrm{λ}$ (iii)

In equation (i), it can be seen that the rate of change of kinetic energy is the function of ${\cos }^2\left(kx-\mathrm{Ó\neg }t\right)$. 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:

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

Similarly, the velocity of the wave is given as:

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

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

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

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

localid="1660983930043" $$\begin{gathered} \frac{dk}{dt}=\frac{1}{2}\mathrm{μ}\mathrm{ν}{\mathrm{Ó\neg }}^2{y}_m^2 \\ 10=\frac{1}{2}\mathrm{μ}\mathrm{ν}{\mathrm{Ó\neg }}^2{y}_m^2 \\ {y}_m^2=\frac{2×10}{2{\mathrm{Ï€}}^2\mathrm{μ}\mathrm{λ}{f}^3}\left(âˆ\mu fromgiven\right) \\ {y}_m^2=\frac{2×10}{\mathrm{μ}\mathrm{ν}{\mathrm{Ó\neg }}^2} \\ {y}_m=\sqrt{\frac{10}{2{\mathrm{Ï€}}^2\mathrm{μ}\mathrm{λ}{f}^3}}\left(âˆ\mu fromequation\left(ii\right)\right) \\ =\sqrt{\frac{10}{2×9.86×0.002×0.20×{500}^3}} \\ =\sqrt{\frac{10}{986000}} \\ =0.0032\mathrm{m} \end{gathered}$$

Hence, the value of the amplitude is 0.0032 m