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JCECE Engineering JCECE Engineering Solved Paper-2003

  • question_answer
    A source and an observer move away from each other, with a velocity of \[20\,\,m/s\]. If the apparent frequency heard by the observer is \[1840\,\,Hz\], the actual frequency of the source is: (Velocity of sound in\[=340\,\,m/s)\]:

    A) \[2486\,\,Hz\]                   

    B) \[2070\,\,Hz\]

    C) \[2134\,\,Hz\]                   

    D) \[1872\,\,Hz\]

    Correct Answer: B

    Solution :

    Key Idea: Movement of the source and observer alters the wavelength and the perceived frequency of sound. As both source and observer are moving away, so observed frequency decreases. \[\therefore \]  \[f'=\left( \frac{v-{{v}_{o}}}{v+{{v}_{s}}} \right)f\] where \[v\] is the velocity of sound\[=340\,\,m/s\]. Here,\[{{v}_{o}}=20\,\,m/s,\,\,{{v}_{s}}=20\,\,m/s,\,\,f'=1840\,\,Hz\] \[\therefore \]  \[1840=\left( \frac{340-20}{340+20} \right)f\] or            \[f=\frac{1840\times 360}{320}=2070\,\,Hz\]


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