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JCECE Engineering JCECE Engineering Solved Paper-2003

  • question_answer
    If the equation of Motion of standing wave is\[y=0.3\sin (314t-1.57x)\], then the velocity of standing wave is:

    A) \[400\,\,unit\]                  

    B) \[350\,\,unit\]

    C) \[209\,\,unit\]                  

    D) \[200\,\,unit\]

    Correct Answer: D

    Solution :

    Key Idea: Standard equation of standing wave is                 \[y=a\sin (\omega t-kx)\] The given equation is                 \[y=0.3\sin (314t-1.57x)\] On comparing the two equations, we have                 \[\omega t=314t\] and        \[kx=1.57x\] \[\Rightarrow \]               \[\omega =314\] and        \[k=1.57\] Hence, velocity of standing wave is                 \[v=\frac{\omega }{k}=\frac{314}{1.57}=200\,\,unit\]


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