Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    The angular velocities of three bodies in simple harmonic motion are\[{{\omega }_{1}},\,\,{{\omega }_{2}},\,\,{{\omega }_{3}}\]with their respective amplitudes as\[{{A}_{1}},\,\,{{A}_{2}},\,\,{{A}_{3}}\]. If all the three bodies have same mass and velocity, then

    A) \[A_{1}^{2}\omega _{1}^{2}=A_{2}^{2}\omega _{2}^{2}=A_{3}^{2}{{\omega }^{2}}\]    

    B) \[A_{1}^{2}{{\omega }_{1}}=A_{2}^{2}{{\omega }_{1}}=A_{3}^{2}{{\omega }_{3}}\]

    C) \[{{A}_{1}}\omega _{1}^{2}={{A}_{2}}\omega _{2}^{2},\,\,{{A}_{3}}\omega _{3}^{2}\]

    D) \[{{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}={{A}_{3}}{{\omega }_{3}}\]

    Correct Answer: D

    Solution :

    Rate of change of displacement is known as velocity. The equation for displacement \[(y)\] of a body in SHM with angular velocity\[\omega \]is given by                 \[y=a\sin \omega t\] where a is amplitude velocity \[v=\] rate of change of displacement\[=\frac{dy}{dt}\]                 \[V=\frac{dy}{dt}=\frac{d}{dt}(a\sin \omega t)=a\omega \cos \omega t\] Given,\[{{v}_{1}}={{v}_{2}}={{v}_{3}}\]and\[{{a}_{1}}={{A}_{1}},\,\,{{a}_{2}}={{A}_{2}}\]                 \[{{a}_{3}}={{A}_{3}}\]                 \[{{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}={{A}_{3}}{{\omega }_{3}}\]


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