1. Given wave equation

   The transverse wave is $y = 2\sin(\omega t-kx)\ \text{cm}$

   Comparing with the standard form $y = A\sin(\omega t-kx),$ we get amplitude $A = 2\ \text{cm}.$

2. Wave velocity

   The wave speed is $v = \frac{\omega}{k}.$

   Also, since $k = \frac{2\pi}{\lambda},$ therefore $v = \frac{\omega}{2\pi/\lambda} = \frac{\omega\lambda}{2\pi}.$

3. Maximum particle velocity

   Particle velocity is the transverse velocity of a medium particle: $v_p = \frac{\partial y}{\partial t} = 2\omega \cos(\omega t-kx).$

   Hence, maximum particle velocity is $v_{p,\max} = 2\omega.$

4. Condition given in the question

   Wave velocity becomes equal to maximum particle velocity: $v = v_{p,\max}.$

   So, $\frac{\omega\lambda}{2\pi} = 2\omega.$

5. Solve for wavelength

   Cancel $\omega$ from both sides: $\frac{\lambda}{2\pi} = 2$ $\lambda = 4\pi\ \text{cm}.$

6. Check options

   • A: $4\pi$ ✅
   • B: $2\pi$ ❌
   • C: $\pi$ ❌
   • D: $2$ ❌

Therefore, the correct answer is A.