Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    Find beat frequency? Motion of two particles is given by \[{{y}_{1}}=0.25\sin (310t)\] \[{{y}_{2}}=0.25\sin (316t)\]

    A)  3                                            

    B) \[\frac{3}{\pi }\]

    C) \[\frac{6}{\pi }\]             

    D)         6

    Correct Answer: B

    Solution :

    The given equations of waves be written as                 \[{{y}_{1}}=0.25\,\,\sin (310\,\,t)\]                          ... (i) and        \[{{y}_{2}}=0.25\,\,\sin (316\,\,t)\]                          ? (ii) Comparing Eqs. (i) and (ii) with the standard wave equation, written as                 \[y=a\,\,\sin (\omega t)\]                                            ... (iii) We have,                 \[{{\omega }_{1}}=310\] \[\Rightarrow \]               \[{{f}_{1}}=\frac{310}{2\pi }\]unit and        \[{{\omega }_{2}}=316\] \[\Rightarrow \]               \[{{f}_{2}}=\frac{316}{2\pi }\]unit Hence, beat frequency                      \[={{f}_{2}}-{{f}_{1}}\]                    =\[\frac{316}{2\pi }-\frac{310}{2\pi }\]                    \[=\frac{3}{\pi }\]unit


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