2. Solved Papers
  3. Manipal Engineering

Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    A train is approaching with velocity 25 m/s towards a pedestrian standing on the track, frequency of horn of train is 1 kHz. Frequency heard by the pedestrain is (v= 350 m/s)

    A)  1077 Hz               

    B)         1167 Hz

    C)  985 Hz                 

    D)         954 Hz

    Correct Answer: A

    Solution :

    Since, train (source) is moving towards pedestrain   (observer),   the   perceived frequency will be higher than the original.                 \[f=f\left( \frac{v+{{v}_{o}}}{v-{{v}_{s}}} \right)\] Here,\[{{v}_{o}}=0\](as observer is stationary) \[{{v}_{s}}=25\,\,m/s\](velocity of source) \[v=350\,\,m/s\](velocity of sound) and \[f=1kHz\](original frequency) Hence,\[f=1000\left( \frac{350+0}{350-25} \right)\]                    \[=1000\times \frac{350}{325}\]                   \[=1077\,\,Hz\]


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