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  4. For the stationary wave y=4 sin (π x/15) cos 96π t The distance between a node and the next antinode is

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Q. For the stationary wave $ y=4\sin \frac{\pi x}{15}\cos 96\pi t $ The distance between a node and the next antinode is

JamiaJamia 2010

Solution:

Given, $ y=4\sin \frac{\pi x}{15}\cos 96\pi t $ For nodes, displacement is zero. $ \therefore $ $ \sin \frac{\pi x}{15}=0 $ $ \Rightarrow $ $ \frac{\pi x}{15}=0,\pi ,2\pi $ $ x=\frac{15}{2},\frac{45}{2}.... $ For antinodes, displacement is maximum. $ \therefore $ $ \sin \frac{\pi x}{15}=1 $ $ \Rightarrow $ $ \frac{\pi x}{15}=\frac{\pi }{2},\frac{3\pi }{2},.... $ $ x=\frac{15}{2},\frac{45}{2}.... $ Hence, distance between a node and next antinode $ =\frac{15}{2}-0=\frac{15}{2}=7.5\,cm $