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AMU Medical AMU Solved Paper-2011

  • question_answer
    The transverse displacement of a string fixed at    both    ends    is    given    by \[y=0.06\sin \left( \frac{2\pi x}{3} \right)\cos (100\pi t)\]where \[x\]and \[y\] are in metre and t is in second. The length of the string is 1.5 m and its mass is \[\text{3}.0\times \text{1}{{0}^{-\text{2}}}\text{ kg}.\]What is die tension in the string?

    A)  225 N                                   

    B)  300 N

    C)  450 N                                   

    D)  675 N

    Correct Answer: D

    Solution :

                     \[y=0.06\sin \left( \frac{2\pi }{3} \right)\cos (100\pi t)\]                 Comparing on standard equation                                 \[y=2a\,\sin \frac{2\pi x}{\lambda }\cos \frac{2\pi vt}{\lambda }\]                 We will get frequency, \[n=\frac{v}{\lambda }=50\]                 Frequency of vibration \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\]                                 \[50=\frac{1}{2\times 1.5}\sqrt{\frac{T}{3\times {{10}^{-2}}}}\]                                 \[50\times 50=\frac{1}{3\times 3}.\,\frac{T}{3\times {{10}^{-2}}}\] or            \[T=50\times 50\times 3\times 3\times 3\times {{10}^{-2}}\]                                 \[T=675\,\,N\]


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