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J & K CET Medical J & K - CET Medical Solved Paper-2009

  • question_answer
    A sound wave with frequency 256 Hz falls normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles will have maximum amplitude of vibrations is nearly ( velocity of sound in air is\[336\text{ }m{{s}^{-1}}\])

    A)  32.8 cm                               

    B)  50 cm

    C)  65.8 cm                               

    D)  25 cm

    Correct Answer: A

    Solution :

                     Given, \[f=256\text{ }Hz\] \[{{v}_{sound}}=336\,m{{s}^{-1}}\]                 \[\lambda =\frac{v}{f}=\frac{336}{256}=1.3125\,m\] \[=131.25\text{ }cm\] The distance between a node\[(N)\]and adjoining antinode\[=\frac{\lambda }{4}=\frac{131.25}{4}=32.8\] cm the shortest distance from the wall at which the air particles will have maximum amplitude of vibration is 32.8 cm.


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