Two progressive waves \(Y_1=\sin 2 \pi\left(\frac{t}{0 \cdot 4}-\frac{x}{4}\right)\) and \(\mathrm{Y}_2=\sin 2 \pi\left(\frac{\mathrm{t}}{0 \cdot 4}+\frac{\mathrm{x}}{4}\right)\) superpose to form a standing wave. ' \(x\) ' and ' \(y\) ' are in SI system. Amplitude of the particle at \(x=0.5 \mathrm{~m}\) is \(\left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)

\(Y=Y_1+Y_2\)
\(\begin{array}{ll}\therefore \quad & Y=\sin \left(\frac{2 \pi t}{0.4}-\frac{2 \pi x}{4}\right)+\sin \left(\frac{2 \pi t}{0.4}+\frac{2 \pi x}{4}\right) \\ & \sin (A+B)+\sin (A-B)=2 \sin A \cos B \\ \therefore \quad & Y=2 \sin \left(\frac{2 \pi t}{0.4}\right), \cos \left(\frac{2 \pi x}{4}\right) \\ & Y=R \sin \left(\frac{2 \pi t}{0.4}\right), \text { where } R \text { is the amplitude } \\ \therefore \quad & R=2 \cos \left(\frac{2 \pi x}{4}\right) \\ \therefore \quad & \text { at } x=0.5 \\ & R=2 \cos \left(\frac{2 \pi}{4 \times 2}\right)=2 \cos \left(\frac{\pi}{4}\right)=\frac{2}{\sqrt{2}}=\sqrt{2}\end{array}\)