A) A displacement node occurs at x = 0.15 m
B) An antinode occurs at x = 0.3 m
C) The wavelength of the wave is 0.2 m
D) The speed of the wave is 5.0 m/s
Correct Answer: A
Solution :
\[y=0.02\cos (10\,\pi x)\cos \,\left( 50\,\pi \,t+\frac{\pi }{2} \right)\] 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 \[\Rightarrow \]\[\cos (10\pi x)=\pm 1\Rightarrow x=0,\pi ,2\pi ...\] Þ x = 0, 0.1m, 0.2m ? Now \[\lambda =2\times \]Distance between two nodes or antinodes = \[2\times 0.1=0.2\]\[m\]and \[\frac{2\pi vt}{\lambda }=50\pi t\] \[v=25\lambda =25\times 0.2\]\[=5m/\sec \].You need to login to perform this action.
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