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AMU Medical AMU Solved Paper-2009

  • question_answer
    The transverse displacement of a string fixed at both      ends      is      given      by \[\text{y}=0.0\text{6 sin}\left( \frac{2\pi x}{3} \right)\text{cos }\left( \text{12}0\text{ }\pi \text{t} \right)\text{y}\]and x are in metres and r in second. The wavelength and frequency of the two superposing waves are

    A)  2 m, 120 Hz       

    B)  2 m, 60 Hz

    C) \[\frac{\text{3}}{2}\text{ m},\text{12}0\text{ Hz}\]         

    D)  3 m, 60 Hz

    Correct Answer: D

    Solution :

                     The general equation of stationary wave is \[y=2a\,\sin kx\,\cos \omega \,t\]                   ... (i) Given equation of stationary wave \[y=0.06\sin \left[ \frac{2\pi \,x}{3} \right]\cos \,[120\,\pi t]\]              ... (ii) Comparing Eqs. (i) and (ii), we get                                 \[k=\frac{2\pi }{3}\]                                 \[\frac{2\pi }{\lambda }=\frac{2\pi }{3}\]               \[\left[ As\,k=\frac{2\pi }{\lambda } \right]\] \[\Rightarrow \]                               \[\lambda =3\,m\] and        \[\omega =120\,\pi \]                 \[2\pi v=120\,\pi \] \[\Rightarrow \]               \[v=60\,Hz\]


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