JCECE Medical JCECE Medical Solved Paper-2005

  • question_answer
    A tuning fork of frequency 580 Hz is employed to produce transverse waves on, a long rope. The distance between the nearest crests is found to be 20 cm. The velocity of the wave is:

    A) \[58\,m{{s}^{-1}}\]

    B)  \[580\,m{{s}^{-1}}\]

    C)  \[20\,m{{s}^{-1}}\]

    D)  \[116\,m{{s}^{-1}}\]

    Correct Answer: D

    Solution :

     Key Idea: Distance between nearest crests is equal to the wavelength. In case of transverse wave, wavelength is the distance between two consecutive crests or throughs. The relation between velocity \[(v),\] frequency  \[(n)\] and wavelength \[(\lambda )\] is \[v=n\lambda \] Given, \[n=580\,Hz,\,\lambda =20\,cm=20\times {{10}^{-2}}m\] \[\therefore \] \[v=580\times 20\times {{10}^{-2}}=116\,m/s\] Note: In case of longitudinal wave, wavelength is the   distance   between   two   successive compressions or rarefactions.


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