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CMC Medical CMC-Medical Ludhiana Solved Paper-2015

  • question_answer
    The sound waves travelling through the air are of wavelength 40 cm. If at the some specified point, the difference between the maximum and minimum pressure is\[1.0\times {{10}^{-3}}N{{m}^{-2}}\] then the amplitude of vibration of medium particles is [The bulk modulus of air is \[1.4\times {{10}^{5}}N{{m}^{-2}}\]]

    A)  \[2.2\times {{10}^{-10}}m\]                       

    B)  \[3.7\times {{10}^{-9}}m\]

    C)  \[1.7\times {{10}^{-11}}m\]                       

    D)  \[1.9\times {{10}^{-10}}m\]

    Correct Answer: A

    Solution :

                    The pressure amplitude is \[{{p}_{0}}=\frac{1.0\times {{10}^{-3}}}{2}=0.5\times {{10}^{-3}}N{{m}^{-2}}\] Now, displacement amplitude. So is given by \[{{p}_{0}}=BK{{S}_{0}}\] \[\Rightarrow \]               \[{{S}_{0}}=\frac{{{p}_{0}}}{BK}=\frac{{{p}_{0}}\lambda }{2\pi B}\]      \[=\frac{0.5\times {{10}^{-3}}\times 40\times {{10}^{-2}}}{2\times 3.14\times 14\times {{10}^{5}}}\]      \[=2.2\times {{10}^{-10}}m\]


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