BHU PMT BHU PMT (Screening) Solved Paper-2009

  • question_answer
    A body executes simple harmonic motion under the action of force\[{{F}_{1}}\]with a time period 4/5 s. If the force is changed to\[{{F}_{2}}\]it executes simple harmonic motion with time period 3/5 s. If both forces\[{{F}_{1}}\]and\[{{F}_{2}}\]act simultaneously in the same direction on the body, its time period will be

    A)  \[\frac{12}{25}s\]                           

    B)  \[\frac{24}{25}s\]

    C)  \[\frac{35}{24}s\]                           

    D)  \[\frac{15}{12}s\]

    Correct Answer: A

    Solution :

    Under the influence of one force\[{{F}_{1}}=m\omega _{1}^{2}y\]and under the action of another force, \[{{F}_{2}}=m\omega _{2}^{2}y\]Under the action of both the forces\[F={{F}_{1}}+{{F}_{2}}\] \[\Rightarrow \]\[m{{\omega }^{2}}y=m\omega _{1}^{2}y+m\omega _{2}^{2}y\] \[\Rightarrow \]\[{{\omega }^{2}}=\omega _{1}^{2}+\omega _{2}^{2}\] \[\Rightarrow \]\[{{\left[ \frac{2\pi }{T} \right]}^{2}}={{\left[ \frac{2\pi }{{{T}_{1}}} \right]}^{2}}+{{\left[ \frac{2\pi }{{{T}_{2}}} \right]}^{2}}\] \[\Rightarrow \]\[T=\sqrt{\frac{T_{1}^{2}\times T_{2}^{2}}{T_{1}^{2}+T_{2}^{2}}}=\sqrt{\frac{{{\left( \frac{4}{5} \right)}^{2}}{{\left( \frac{3}{5} \right)}^{2}}}{{{\left( \frac{4}{5} \right)}^{2}}+{{\left( \frac{3}{5} \right)}^{2}}}}=\frac{12}{25}s\]


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