JCECE Medical JCECE Medical Solved Paper-2004

  • question_answer
    A   transverse   wave   is   given   by \[y=A\,\sin 2\pi (ft-x/\lambda ).\]The maximum particle velocity is 4 times the wave velocity when:

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

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

    C)  \[\lambda =\pi A/2\]

    D) \[\lambda =\pi A/4\]

    Correct Answer: C

    Solution :

     The equation of given wave travelling with amplitude A is given by \[y=A\sin 2\pi \left( ft-\frac{x}{\lambda } \right)\] Maximum velocity\[{{v}_{m}}=A\omega =A\times 2\pi f\] Also,      \[c=f\lambda \] \[\therefore \] \[{{v}_{m}}=\frac{A2\pi c}{\lambda }\] Given, \[{{v}_{m}}=4c,\]then \[4c=\frac{A.2\pi c}{\lambda }\] \[\Rightarrow \] \[\lambda =\frac{\pi A}{2}\]


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