JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is \[Y=0.3\sin \]\[(0.157x)cos\]\[(200\pi t)\]. The length of the string is : (All quantities are in SI units.)             [JEE Main 9-4-2019 Morning]

    A) 20 m               

    B) 80 m

    C) 60 m               

    D) 40 m

    Correct Answer: B

    Solution :

    4th harmonic \[4\frac{\lambda }{2}=\ell \]                                      \[2\lambda =\ell \] From equation\[\frac{2\pi }{\lambda }=0.157\] \[\lambda =40\] \[\ell =2\lambda \] \[=80m\] Option [b]


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