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MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    Two sound waves with wavelengths 4 m and 5 m respectively each propagating in a gas with velocity 340 m/s. The number of beats per second will be

    A)  12 Hz                                   

    B)  15Hz

    C)  17Hz                                    

    D)  11 Hz

    Correct Answer: C

    Solution :

                    As frequency \[=\frac{velocity}{wavelength}\] So,          \[{{f}_{1}}=\frac{v}{{{\lambda }_{1}}}=\frac{340}{4}=85\,Hz\] and        \[{{f}_{2}}=\frac{v}{{{\lambda }_{2}}}=\frac{340}{5}=68\,Hz\] The number of beats per second is the beat frequency and given by                 \[\Delta {{f}_{b}}={{f}_{1}}-{{f}_{2}}=85-68=17\,Hz\]


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