A transverse wave is represented by $y=y_0\sin \left\{\frac{2\text{π}}{\lambda }\left(vt-x\right)\right\}$. For what value of  $\lambda$ is the maximum particle velocity equal to twice the wave velocity?

Given  $y=y_0\sin \left\{\frac{2\text{π}}{\lambda }\left(vt-x\right)\right\}$
$\omega =\frac{2\text{π}}{\lambda }v,k=\frac{2\text{π}}{\lambda }$
Velocity of wave $V_{\omega }=v$
Velocity of particle  $V_P=\frac{\partial y}{\partial t}=y_0\frac{2\text{π}v}{\lambda }\cos \left\{\frac{2\text{π}}{\lambda }(vt-x)\right\}$
Maximum velocity of particle  $V_{\max }=\frac{2\text{π}v{y}_0}{\lambda }$
given that  $V_{\max }=2V_{wave}$
$\therefore \frac{2\text{π}v{y}_0}{\lambda }=2v$
$\therefore \lambda =\text{π}{y}_0$
Therefore, the correct answer is ( D )