The equation of a wave on a string of linear mass density 0.04 kg m^-1 is given by

$y=0.02\left(m\right) \sin \left[2\mathrm{\pi }\left(\frac{\mathrm{t}}{0.04\left(\mathrm{s}\right)}-\frac{\mathrm{x}}{0.50\left(\mathrm{m}\right)}\right)\right].$

The tension in the string is

$$\begin{gathered} y=0.02\left(m\right) \sin \left[2\mathrm{\pi }\left(\frac{\mathrm{t}}{0.04\left(\mathrm{s}\right)}-\frac{\mathrm{x}}{0.50\left(\mathrm{m}\right)}\right)\right] \\ \mathrm{But} y=a \sin (\omega t-kx) \\ \therefore \omega =\frac{2\mathrm{\pi }}{0.04} \Rightarrow \mathcal{v}=\frac{1}{0.04}=25Hz \\ k=\frac{2\mathrm{\pi }}{0.50} \Rightarrow \lambda =0.5 m \\ \therefore velocity, v=\mathcal{v}\lambda =25\times 0.5 \mathrm{m}/\mathrm{s}=12.5 \mathrm{m}/\mathrm{s} \end{gathered}$$

Velocity on a string is given by

            $v=\sqrt{\frac{T}{\mu }} \therefore T=v^2\times \mu ={\left(12.5\right)}^2\times 0.04=6.25N$