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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    A source of sound is moving with constant velocity of 20 m/s emitting a note of frequency 1000 Hz. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be (speed of sound v = 340 m/s)

    A)  9 : 8                                      

    B)  8 : 9

    C)  1:1                                        

    D)  9 :10

    Correct Answer: A

    Solution :

    Apparent frequency when source is coming closer to observer \[{{v}_{c}}=\left( \frac{v}{v-{{v}_{s}}} \right)v\]                   ?.(i) Apparent frequency when source is moving away from observer                 \[{{v}_{f}}=\left( \frac{v}{v+{{v}_{s}}} \right)v\]                 ?..(ii) From two equations                 \[\frac{{{v}_{c}}}{{{v}_{f}}}=\frac{v+{{v}_{s}}}{v-{{v}_{s}}}\]                 \[=\frac{340+20}{340-20}=\frac{9}{8}\] So,          \[{{v}_{c}}:{{v}_{f}}=9:8\]


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