BHU PMT BHU PMT (Mains) Solved Paper-2009

  • question_answer

    Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:

    If two orthogonal SHM of same frequency having initial phase difference of n/2 acts simultaneously on a particle free to move, the particle can move in a (1) straight line                  (2) circle (3) parabola                        (4) ellipse

    A)  1, 2 and 3 are correct

    B)  1 and 2 are correct

    C)   2 and 4 are correct

    D)  1 and 3 are correct

    Correct Answer: C

    Solution :

                     \[x=a\sin \omega t\]and \[y=b\sin \left( \omega t+\frac{\pi }{2} \right)\]                 \[=b\cos \omega t\]                 \[\therefore \]\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}={{\sin }^{2}}\omega t+{{\cos }^{2}}\omega t=1\]                 It is an equation of ellipse, if\[a=b,\]then                 \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]or \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]                 It is an equation of circle.


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