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J & K CET Medical J & K - CET Medical Solved Paper-2003

  • question_answer
    An object producing a pitch of 400Hz flies past a stationary person. The object was moving in a straight line with a velocity 200m/s. The velocity of sound is 300 m/s. What is the change in frequency noted by the person as the object flies past him?

    A)  1440 Hz                                               

    B)  240 Hz

    C)  1200 Hz                                               

    D)  960 Hz

    Correct Answer: C

    Solution :

                    From Dopplers effect, the perceived frequency\[(f)\]is given by \[f=\left( \frac{v-{{v}_{o}}}{v-{{v}_{s}}} \right)f\] where\[{{v}_{o}}\]is velocity of observer,\[{{v}_{s}}\]. of source, \[v\]of sound and\[f\]the original frequency. Given,\[{{v}_{o}}=0\] (stationary), \[v=300\text{ }m/s\] \[{{v}_{s}}=200m/s,\text{ }f=400Hz.\] \[\therefore \]  \[f=\frac{300\times 400}{300-200}=\frac{300\times 400}{100}\] \[\Rightarrow \]               \[f=1200\text{ }Hz\]


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