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JEE Main & Advanced Physics Wave Mechanics Question Bank Critical Thinking

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    A wave disturbance in a medium is described by \[y(x,\,t)=0.02\cos \,\left( 50\,\pi t+\frac{\pi }{2} \right)\cos (10\pi x)\], where x and y are in metres and t in seconds                                            [IIT 1995]

    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 \[\frac{2\pi vt}{\lambda }=50\pi t\]                    \[v=25\lambda =25\times 0.2\]\[=5m/\sec \].


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