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MGIMS WARDHA MGIMS WARDHA Solved Paper-2012

  • question_answer
    Equation of a progressive sound wave is\[y=a\sin \left( 400\pi t-\frac{\pi x}{0.85} \right)\]where\[x\]in (metre)\[t\] (second), then frequency of wave is

    A)  200 Hz                                 

    B)  400 Hz

    C)  500 Hz                                 

    D)  600 Hz

    Correct Answer: A

    Solution :

                    Equation of wave is \[=a\sin \left( 400\pi t-\frac{\pi x}{0.85} \right)\] Comparing this equation with                 \[y=a\sin \left( \omega t-\frac{2\pi x}{\lambda } \right)\]                 \[\omega =400\pi \] \[\therefore \]  \[2\pi n=400\pi \] \[n=200\text{ }Hz\]


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