A transverse wave is represented by $$y = 2\sin(\omega t - kx) \text{ cm}$$. The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity, will be

The transverse wave is $$y = 2\sin(\omega t - kx) \text{ cm}$$ and we need to find the wavelength for which wave velocity equals maximum particle velocity.

The amplitude is $$A = 2 \text{ cm}$$ and the wave velocity is $$v = \dfrac{\omega}{k}$$.

Since the particle velocity is $$\dfrac{\partial y}{\partial t} = 2\omega \cos(\omega t - kx)$$, its maximum value is $$v_{max} = A\omega = 2\omega$$.

Equating the wave velocity to the maximum particle velocity gives $$\dfrac{\omega}{k} = 2\omega$$, which simplifies to $$\dfrac{1}{k} = 2$$ and hence $$k = \dfrac{1}{2}$$.

Since $$k = \dfrac{2\pi}{\lambda}$$, we have $$\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{1/2} = 4\pi \text{ cm}$$.

The correct answer is Option A: $$4\pi$$.