\(y=0.2 \sin 0.8 x \cos 3000 t\)
The distance between any two points of minimum or maximum displacement is simply wavelength \(\left(\frac{\lambda}{4}\right)\) of the wave.
\(k=\frac{2 \pi}{\lambda}=0.8\)
\(\text { or } \lambda=\frac{2 \pi}{0.8}\)
\(\therefore \lambda=\frac{5 \pi}{2} m \Rightarrow \frac{\lambda}{4}=\frac{5 \pi}{8} \,m\)
Hence answer must be \((d)\).
${y}=1.0\, {mm} \cos \left(1.57 \,{cm}^{-1}\right) {x} \sin \left(78.5\, {s}^{-1}\right) {t}$
${x}>0$ ના ક્ષેત્રમાં ઉગમબિંદુથી નજીકનું નિસ્પંદ બિંદુ ${x}=\ldots \ldots \ldots\, {cm}$ અંતરે હશે.