A plane wave (y=a (ω t-k x)) propagates through a stretched string....

Plane wave is given as (y=a \sin (\omega t-k x)) at $t=0$, $$\begin{aligned} & y=a \sin (\omega \times 0-k x) \\ & \Rightarrow \quad=a \sin (-k x) \Rightarrow-a \sin k x \\ \end{aligned}$$) Particle velocity, $$\begin{aligned} v_{\mathrm{pa}} & =\frac{d y}{d t}=\frac{d}{d t}(-a \sin k x) \\ & =-a k \cos k x=-a k \cos\frac{2 \pi}{\lambda} x\left[\because k=\frac{2 \pi}{\lambda}\right] \\ & =-\frac{2 \pi a}{\lambda}\cos\frac{2 \pi x}{\lambda} \quad \ldots (i) \end{aligned}$$)$From$Eq. (i), When (x=0, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cdot \cos 0=\frac{-2 \pi a}{\lambda}) When (x=\frac{\lambda}{4}, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cdot \cos \frac{\pi}{2}=0) When (x=\frac{\lambda}{2}, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cdot \cos \pi=\frac{2 \pi a}{\lambda}) When (x=\frac{3 \lambda}{4}, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cos \left(\frac{-\pi}{2}\right)=0) When (x=\lambda, v_{\mathrm{pa}}=\frac{-2 \pi a}{\lambda} \cdot \cos 2 \pi=\frac{-2 \pi a}{\lambda}) Hence, correct graph is