Q117 / 150 · MHT CET 2007 · Physics

If a source emitting waves of frequency $f$ moves towards an observer with a velocity $\frac{v}{4}$ and the observer moves away from the source with a velocity $v / 6$, the apparent frequency as heard by the observer will be $(v=$ velocity of sound)

  • A
    $\frac{14}{15} f$
  • B
    $\frac{14}{9} f$
  • Correct answer
    C
    $\frac{10}{9} f$
  • D
    $\frac{2}{3} f$

Solution

When source and observer both are moving in the same direction and observer is ahead of source, then apparent frequency $$ f^{\prime}=\frac{v-v_{o}}{v-v_{s}} f $$ $$ =\frac{v-\frac{v}{6}}{v-\frac{v}{4}}=\frac{\frac{5 v}{6}}{\frac{3 v}{4}} f=\frac{10}{9} f $$

Hence, the correct answer is option: 'C'.

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