A transverse wave is represented by the equation, $\mathrm{y}={\mathrm{y}}_0\sin \frac{2\mathrm{\pi }}{\mathrm{\lambda }}(\mathrm{vt}-\mathrm{x})$. For what value of $\mathrm{\lambda }$, the maximum particle velocity is equal to two times the wave velocity ?

On comparing the given equation with standard equation $\mathrm{y}=\mathrm{asin}\frac{2\mathrm{\pi }}{\mathrm{\lambda }}(\mathrm{vt}-\mathrm{x})$ ,

It is clear that wave speed, ${\left(\mathrm{v}\right)}_{\mathrm{wave}}=\mathrm{v}$  ,and ,  
maximum particle velocity, ${\left({\mathrm{v}}_{max}\right)}_{\text{particle }}=\mathrm{a\omega }={\mathrm{y}}_0\times \frac{2\mathrm{\pi v}}{\mathrm{\lambda }}$
$\because {\left({\mathrm{v}}_{max}\right)}_{\text{particie }}=2(\mathrm{v}{)}_{\text{wane }}\Rightarrow \mathrm{\lambda }={\mathrm{\pi y}}_0$