Physics-
General
Easy

Question

An object producing a pitch of 400 Hz flies past a stationary person. The object was moving in a straight line with a velocity200 blank m s to the power of negative 1 end exponent. What is the change in frequency noted by the person as the person as the object files past him?

  1. 1440 Hz    
  2. 240 Hz    
  3. 1200 Hz    
  4. 960 Hz    

The correct answer is: 1200 Hz


    From Doppler’s effect, the perceived frequency (v’) is given by
    v to the power of ´ end exponent equals open parentheses fraction numerator v minus v subscript o end subscript over denominator v minus v subscript s end subscript end fraction close parentheses v
    Where v subscript o end subscript is velocity of observer, v subscript s end subscript of source, v of sound and v the original frequency.
    Given, v subscript o end subscript equals 0(stationary), v equals 300 blank m s to the power of negative 1 end exponent
    v subscript s end subscript equals 200 blank m s to the power of negative 1 end exponent comma blank v equals 400 blank H z
    therefore v to the power of ´ end exponent equals fraction numerator 300 cross times 400 over denominator 300 minus 200 end fraction
    equals fraction numerator 300 cross times 400 over denominator 100 end fraction
    ⟹ v to the power of ´ end exponent equals 1200 blank H z

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