2. Solved Papers
  3. AMU Medical

AMU Medical AMU Solved Paper-2013

  • question_answer
    The transverse displacement of a string fixed  at both ends is given by\[y=0.6\sin \left( \frac{2\pi x}{3} \right)\cos \,(120\,\pi t)\] where \[x\]and y are in metres and t is in seconds. The length of the string is 1.5 m and its mass is\[3.0\times {{10}^{-2}}kg\]. The tension in the string is equal to

    A)  648 N

    B)  724 N

    C)  832 N                   

    D)  980 N

    Correct Answer: A

    Solution :

                     Given that                 \[y=0.06\sin \left( \frac{2\pi x}{3} \right)\cos (120\,\pi t)\] This is comparing with                 \[y=2a\sin \left( \frac{2\pi x}{\lambda } \right)\cos \left( \frac{2\,\pi vt}{T} \right)\] we get                 \[\lambda =3\,m,\,\,v=60\,Hz\] Now      \[v=v\lambda \]                 \[=60\times 3\]                 = 180 m/s Now      \[v=\sqrt{\left( \frac{T}{\mu } \right)}\]                 \[180=\sqrt{\left( \frac{T}{\mu } \right)}\]                 \[T=-180\times 180\times \mu \]                 \[=180\times 180\times \frac{m}{l}\]                 \[=180\times 180\times \frac{3\times {{10}^{-2}}}{1.5}\]                 = 648 N


You need to login to perform this action.
You will be redirected in 3 sec spinner