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AMU Medical AMU Solved Paper-1996

  • question_answer
    Transverse wave traveling on a rope is expressed by the equation \[y=10\sin (4\pi t-0.02\pi x)\]where\[x,\]y are in cms and\[t\]is in second. The wavelength and the maximum velocity of a particle in the rope are given in cm and cm/s, respectively by

    A)  \[50,\text{ }40\pi \]          

    B)  \[100,\text{ 4}0\pi \]

    C)  \[100,\text{ 2}0\pi \]      

    D)  \[100,\text{ 8}0\pi \]

    Correct Answer: B

    Solution :

    : \[y=10sin(4\pi t-0.02\pi x)\] Standard equation is given by \[y=a\sin \left( \frac{2\pi vt}{\lambda }-\frac{2\pi }{\lambda }x \right)=a\sin (\omega t-kx)\] \[\therefore \] \[a=10\]cm                                 ...(i) \[\frac{2\pi }{\lambda }=0.02\pi \] or \[\lambda =100\]cm                            ...(ii) \[\omega =4\pi \] \[\therefore \]Maximum velocity\[=a\omega =10\times 4\pi \] \[=40\pi \]cm/sec        ...(iii) Hence, from (ii) and (iii), we have \[\lambda =100\text{ }cm\] \[{{v}_{m}}=100\pi \] cm/sec.


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