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