A string of length I is fixed at both ends and is vibrating in second harmonic. The amplitude at antinode is 2 mm. The amplitude of a particle at a distance $\frac{\ell }{8}$ from the fixed end is

In a vibrating string, the amplitude at an antinode (point of maximum displacement) is the maximum amplitude observed in the standing wave pattern. For the second harmonic, there are two antinodes and one node (point of zero displacement) between them.

I=λ; 2A= Amplitude at Antinode =2 mm
Amplitude =2Asin [kx]
$\begin{array}{l}=2\sin \left[\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\left(\frac{\mathrm{\ell }}{8}\right)\right]\mathrm{mm}\\ =2\sin \left(\frac{\mathrm{\pi }}{\mathrm{\lambda }}\frac{\mathrm{\lambda }}{4}\right)\mathrm{mm}=2\sin \left(\frac{\mathrm{\pi }}{4}\right)\mathrm{mm}\\ =\frac{2\mathrm{mm}}{\sqrt{2}}=\sqrt{2}\mathrm{mm}\end{array}$