CET Karnataka Medical CET - Karnataka Medical Solved Paper-2013

  • question_answer
    A person with vibrating tuning fork of frequency 338 Hz is moving towards a vertical wall with a speed of \[2\text{ }m{{s}^{-1}}\]. Velocity of sound in air is \[340m{{s}^{-1}}\]. The number of beats heard by that person per second is

    A)  \[2\]     

    B)  \[4\]    

    C)  \[6\]     

    D)  \[8\]

    Correct Answer: B

    Solution :

    : Frequency of tuning fork (source), \[\upsilon =338Hz\] Velocity of sound, \[\upsilon =340\text{ }m{{s}^{-1}}\] When the person is moving towards the wall, the frequency of sound reaching the wall is \[\upsilon =\frac{\upsilon \upsilon }{\upsilon -{{\upsilon }_{s}}}\]                ??.(i) On reflection wall acts as source and person is the listener moving towards the source. So the frequency heard by the person after reflection from the wall is \[\upsilon =\left[ \frac{\upsilon +{{\upsilon }_{L}}}{\upsilon -{{\upsilon }_{S}}} \right]\upsilon \]         ??..(ii) Substituting the value of v from Eq. (i) in (ii), we get \[\upsilon =\left[ \frac{\upsilon +{{\upsilon }_{L}}}{\upsilon -{{\upsilon }_{S}}} \right]\upsilon =\left( \frac{\upsilon +{{\upsilon }_{S}}}{\upsilon -{{\upsilon }_{S}}} \right)\upsilon \]   [As \[{{\upsilon }_{L}}={{\upsilon }_{S}}\]] \[=\left[ \frac{340+2}{340-2} \right]338=\frac{342}{338}\times 338=342Hz\] \[\therefore \]Number of beats heard per second \[=\upsilon -\upsilon \] \[=342-338=4\]


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