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JEE Main & Advanced Physics Wave Mechanics Question Bank Stationary Waves

  • question_answer
    The equation of stationary wave along a stretched string is given by \[y=5\sin \frac{\pi x}{3}\cos 40\pi t\], where x and y are in cm and t in second. The separation between two adjacent nodes is         [CPMT 1990; MP PET 1999; AMU 1999; DPMT 2004; BHU 2005]

    A)            1.5 cm                                     

    B)            3 cm

    C)            6 cm                                         

    D)            4 cm

    Correct Answer: B

    Solution :

                         On comparing the given equation with standard equation \[y=2a\sin \frac{2\pi x}{\lambda }\cos \frac{2\pi vt}{\lambda }\]Þ\[\frac{2\pi x}{\lambda }=\frac{\pi x}{3}\Rightarrow \lambda =6\]            Separation between two adjacent nodes = \[\frac{\lambda }{2}=3\]cm


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