d
(d) \(y = 0.02\cos (10\,\pi x)\cos \,\left( {50\,\pi \,t + \frac{\pi }{2}} \right)\)
At node, amplitude \(= 0\)
\( \Rightarrow \) \(\cos (10\pi x) = 0 \Rightarrow 10\,\pi x = \frac{\pi }{2},\frac{{3\pi }}{2}\)
\( \Rightarrow \,x = \frac{1}{{20}} = 0.05 m, 0.15m …..\)
At antinode, amplitude is maximum
\( \Rightarrow \)\(\cos (10\pi x) = \pm 1 \Rightarrow x = 0,\pi ,2\pi ...\)
\(==> x = 0, 0.1m, 0.2m …\)
Now \(\lambda = 2 \times \) Distance between two nodes or antinodes
= \(2 \times 0.1 = 0.2 m\) and \(\frac{{2\pi vt}}{\lambda } = 50\pi t\)
\(v = 25\lambda = 25 \times 0.2  = 5m/\sec \).