A transverse wave is represented by: $$y = \frac{10}{\pi} \sin\left(\frac{2\pi}{T}t - \frac{2\pi}{\lambda}x\right)$$. For what value of the wavelength the wave velocity is twice the maximum particle velocity?

From the wave equation $$y = \frac{10}{\pi} \sin \left( \frac{2\pi}{T}t - \frac{2\pi}{\lambda}x \right)$$, amplitude $$A = \frac{10}{\pi}$$ and the wave velocity $$v_w = \frac{\omega}{k} = \frac{\lambda}{T}$$.

The maximum particle velocity is given by $$v_{p(max)} = A\omega = \left( \frac{10}{\pi} \right) \left( \frac{2\pi}{T} \right) = \frac{20}{T}$$.
Substituting these into the given condition $$v_w = 2 v_{p(max)}$$ yields $$\frac{\lambda}{T} = 2 \times \frac{20}{T}$$, which simplifies to $$\lambda = 40$$ cm.