JEE Main & Advanced Physics Wave Mechanics Question Bank Stationary Waves

  • question_answer
    For the stationary wave \[y=4\sin \,\left( \frac{\pi x}{15} \right)\cos (96\,\pi t)\], the distance between a node and the next antinode is[MP PMT 1987]

    A)            7.5 

    B)            15

    C)            22.5                                          

    D)            30

    Correct Answer: A

    Solution :

                         Comparing given equation with standard equation                    \[y=2a\sin \frac{2\pi x}{\lambda }\cos \frac{2\pi vt}{\lambda }\]gives us \[\frac{2\pi }{\lambda }=\frac{\pi }{15}\Rightarrow \lambda =30\] Distance between nearest node and antinodes = \[\frac{\lambda }{4}=\frac{30}{4}=7.5\]


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