JCECE Engineering JCECE Engineering 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 through. 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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