But λ = 2π / k , hence—
vk = 2πC / v
Hence if we differentiate with respect to v, we have—
d(vk) / dv = – 2πC / v²
Again, k = 2π / λ = 2πC / v² ; therefore—
d(k) / dv = – 2 2πC / v³
Hence, dividing the expression for d(vk) / dv by that for d(k) / dv , we have—
V = d(vk) / d(k) = v / 2
In other words, the wave-train velocity is equal to half the wave-velocity. This is the case with deep-sea waves. Suppose, however, that, as in the case of air waves, the wave-velocity is independent of the wave-length. Then if two trains of waves of slightly different wave-length are superposed, we have k and k′ different in value but nearly equal, and v and v′ equal. Hence the equation (i.) takes the form—
V = v