Given,  $y=a\sin 2\pi (bt-cx)$

On comparing this equation with general equation

$y=r\sin \left(\frac{2\pi t}{T}-\frac{2\pi }{\lambda }x\right),\text{ we get }$

$\frac{2\pi }{T}=\omega =2\pi b,r=a$

$\lambda =\frac{1}{c}\text{ and }T=\frac{1}{b}$

Maximum particle velocity,  $\omega r=2\pi ba$

Wave velocity,  $v=\frac{\lambda }{T}=\frac{b}{c}$

Given, maximum particle velocity = 2 x wave velocity

$2\pi ba=2\times \frac{b}{c}\Rightarrow c=\frac{1}{\pi a}$