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JCECE Engineering JCECE Engineering Solved Paper-2002

  • question_answer
    A transverse wave is given by\[y=A\sin 2\pi \left( \frac{t}{T}-\frac{x}{\lambda } \right)\]. The maximum particle, velocity is equal to 4 times the wave velocity, when:

    A) \[\lambda =2\pi A\]                       

    B) \[\lambda =\frac{1}{2}\pi A\]

    C)  \[\lambda =\pi \,\,A\]                  

    D)  \[\lambda =\frac{1}{4}\pi A\]

    Correct Answer: B

    Solution :

    The given equation of transverse wave is                 \[y=A\sin 2\pi \left( \frac{t}{T}-\frac{x}{\lambda } \right)\] Maximum particle velocity\[({{u}_{m}})=a\omega \] From standard wave equation,\[a=A\],                 \[\omega =2\pi n=2\pi \frac{v}{\lambda }\] \[\therefore \]  \[{{u}_{m}}=A\cdot 2\pi \frac{v}{\lambda }=4v\] \[\Rightarrow \]               \[\lambda =\frac{\pi A}{2}\]


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