A wave disturbance in a medium is described by  $\mathrm{y}(\mathrm{x},\mathrm{t})=0.02\cos \left(50\mathrm{\pi t}+\frac{\mathrm{\pi }}{2}\right)\cos \left(10\mathrm{\pi x}\right)$, where x and y are in metres and t in seconds

$\mathrm{y}=0.02\cos \left(10\mathrm{\hspace{0.17em}}\mathrm{\pi x}\right)\cos \mathrm{\hspace{0.17em}}\left(50\mathrm{\hspace{0.17em}}\mathrm{\pi }\mathrm{\hspace{0.17em}}\mathrm{t}+\frac{\mathrm{\pi }}{2}\right)$

At node, amplitude = 0

$\Rightarrow \cos \left(10\mathrm{\pi x}\right)=0\Rightarrow 10\mathrm{\pi x}=\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}$

$\Rightarrow \mathrm{x}=\frac{1}{20}=0.05\mathrm{m}, 0.15\mathrm{m} ......$

At antinode, amplitude is maximum 

$\Rightarrow \cos \left(10\mathrm{\pi x}\right)=\pm 1\Rightarrow \mathrm{x}=0,\mathrm{\pi },2\mathrm{\pi }...$

$\Rightarrow$x = 0, 0.1m, 0.2m … 

Now $\mathrm{\lambda }=2\times$Distance between two nodes or antinodes 

$=2\times 0.1=0.2\mathrm{m} \mathrm{and} \frac{2\mathrm{\pi vt}}{\mathrm{\lambda }}=50\mathrm{\pi t}$

$\mathrm{v}=25\mathrm{\lambda }=25\times 0.2=5\mathrm{m}/\sec .$