2. Solved Papers
  3. J & K CET Engineering

J & K CET Engineering J and K - CET Engineering Solved Paper-2003

  • question_answer
    An object producing a pitch of \[400\text{ }Hz\]flies past a stationary person. The object was moving in a straight line with a velocity 200 m/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)  \[1440\text{ }Hz\]

    B)  \[240\text{ }Hz\]

    C)  \[1200\text{ }Hz\]     

    D)  \[960\text{ }Hz\]

    Correct Answer: C

    Solution :

    From Doppler's 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 / the original frequency. Given,  \[{{v}_{o}}=0\]  (stationary), \[v=300\text{ }m/s\] \[{{v}_{s}}=200m/s,\,\,\,\,\,\,f=400Hz.\] \[\therefore \] \[f'=\frac{300\times 400}{300-200}=\frac{300\times 400}{100}\] \[\Rightarrow \] \[f'=1200Hz\]


You need to login to perform this action.
You will be redirected in 3 sec spinner