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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1999

  • 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 \[y=a\sin \left( \omega t-\frac{2\pi x}{\lambda } \right)\] \[\omega =400\pi \] \[\therefore \,2\pi n=400\pi \] \[n=200Hz\]


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