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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2010

  • question_answer
        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

    A)  7.5cm                                  

    B)  15cm

    C)  22.5cm                                

    D)  30cm

    Correct Answer: A

    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                 \[=\frac{15}{2}-0=\frac{15}{2}=7.5\,cm\]


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