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EAMCET Medical EAMCET Medical Solved Paper-1997

  • question_answer
    A source and an observer move away from each other, each with a velocity of 10 m/s, with respect to ground. If the observer finds the frequency of sound coming from the source as 1950 Hz, the original frequency of the source is : (assume, velocity of sound in air = 340 m/s)

    A)  1950 Hz                               

    B)  2068 Hz

    C)  1832 Hz                                               

    D)  2186 Hz

    Correct Answer: B

    Solution :

                     Applying Dopplers effect, if source and observer both are moving away from each other. Then, the apparent frequency heard by observer is given by \[n=\left( \frac{v-{{v}_{o}}}{v+{{v}_{s}}} \right)n\]    \[\left( \begin{align}   & Given:n=1950\,Hz \\  & {{v}_{o}}={{v}_{s}}=10\,m/s \\  & v=340\,m/s \\ \end{align} \right)\] \[1950=\left( \frac{340-10}{340+10} \right)n\]                 \[\Rightarrow \]               \[n=\frac{1950}{\frac{340-10}{340+10}}=\frac{1950\times 350}{330}\]                                 \[=2068\,Hz\]


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