When a wave travels in a medium, the particle displacement is given by : y = a sin 2π(bt – cx) ; where a, b and c are constants. The maximum particle velocity will be twice the wave velocity if: c = πa    b = acb =

The given wave equation is y = a sin 2π(bt – cx).

The particle velocity in a wave is given by the derivative of the displacement with respect to time. So, we differentiate the given equation with respect to time (t).

dy/dt = a * 2πb * cos(2π(bt - cx))

The maximum value of cos(2π(bt - cx)) is 1. So, the maximum particle velocity (Vp max) is a * 2πb.

The wave velocity (Vw) is given by the ratio of the frequency to the wavelength, which in this case is b/c (from the wave equation).

The problem states that the maximum particle velocity is twice the wave velocity. So, we can set up the equation:

Vp max = 2 * Vw

Substituting the expressions we found for Vp max and Vw, we get:

a * 2πb = 2 * (b/c)

Solving this equation for c, we get:

c = πa

The problem also asks for the value of b. Since we know that c = πa, we can substitute this into the equation for the wave velocity:

b = c * Vw = πa * (b/c) = πa * (b/πa) = b

So, b = acb = a * πa * b = πa^2b

Therefore, the conditions for the maximum particle velocity to be twice the wave velocity are c = πa and b = acb = πa^2b.