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JEE Main & Advanced Physics Wave Mechanics Question Bank Basics of Mechanical Waves

  • question_answer
    A wave of frequency 500 Hz has velocity 360 m/sec. The distance between two nearest points 60° out of phase, is                   [NCERT 1979; MP PET 1989; JIPMER 1997;  RPMT 2002, 03; CPMT 1979, 90, 2003; BCECE 2005]

    A)            0.6 cm                                     

    B)            12 cm

    C)            60 cm                                      

    D)            120 cm

    Correct Answer: B

    Solution :

               The distance between two points i.e. path difference between them \[\Delta =\frac{\lambda }{2\pi }\times \varphi =\frac{\lambda }{2\pi }\times \frac{\pi }{3}=\frac{\lambda }{6}=\frac{v}{6n}\] \[(\,\because \,\,v=n\lambda )\] Þ \[\Delta =\frac{360}{6\times 500}=0.12\,m=12\,cm\]


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