JEE Main & Advanced Physics Wave Mechanics Question Bank Beats

  • question_answer
    The frequencies of two sound sources are 256 Hz and 260 Hz. At t = 0, the intensity of sound is maximum. Then the phase difference at the time t = 1/16 sec will be

    A)            Zero                                         

    B)            p

    C)            p/2

    D)            p/4

    Correct Answer: C

    Solution :

               Time interval between two consecutive beats                    \[T=\frac{1}{{{n}_{1}}-{{n}_{2}}}=\frac{1}{260-256}=\frac{1}{4}\sec \] so, \[t=\frac{1}{16}=\frac{T}{4}\,\,\sec \]                    By using time difference =\[\frac{T}{2\pi }\times \]Phase difference                    \[\Rightarrow \]\[\frac{T}{4}=\frac{T}{2\pi }\times \varphi \Rightarrow \varphi =\frac{\pi }{2}\]


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