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JEE Main & Advanced Physics Wave Mechanics Question Bank Beats

  • question_answer
    Two vibrating tuning forks produce progressive waves given by \[{{Y}_{1}}=4\sin 500\pi t\]and \[{{Y}_{2}}=2\sin 506\pi t.\] Number of beats produced per minute is                                                     [CBSE PMT 2005]

    A)            360

    B)            180

    C)            3    

    D)            60

    Correct Answer: B

    Solution :

               From the given equations of progressive waves \[{{\omega }_{1}}=500\pi \] and \[{{\omega }_{2}}=506\pi \] \ \[{{n}_{1}}=250\]  and \[{{n}_{2}}=253\]            So beat frequency \[={{n}_{2}}-{{n}_{1}}=253-250=3\] beats per sec \ Number of beats per min = 180.


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