A) Circular
B) Straight line
C) Parabolic
D) Elliptical
Correct Answer: D
Solution :
Let two perpendicular simple harmonic motions are \[x=A\,\sin \left( \omega t \right)\] ?(1) \[y=B\,\sin \left( \omega t+\phi \right)\] ?(2) From Eq. (1) \[\sin \,\left( \omega t \right)=\frac{x}{A}\] \[\Rightarrow \] \[\cos \,\omega t=\sqrt{1-\frac{{{x}^{2}}}{{{A}^{2}}}}\] From Eq. (2)\[\frac{y}{B}=\sin \left( \omega t \right)\cos \phi +\cos \left( \omega t \right)\,\sin \,\phi \] \[=\left( \frac{x}{A} \right)\cos \,\,\phi +\sqrt{1-\frac{{{x}^{2}}}{{{A}^{2}}}}\sin \,\,\phi \] \[\Rightarrow \] \[\frac{y}{B}-\frac{x}{A}\cos \,\,\phi =\sqrt{1-\frac{{{x}^{2}}}{{{A}^{2}}}}\sin \,\,\phi \] On squaring, we get \[=\frac{{{y}^{2}}}{{{B}^{2}}}+\frac{{{x}^{2}}}{{{A}^{2}}}{{\cos }^{2}}\,\,\phi -\frac{2xy}{AB}\cos \,\,\phi =\left( 1-\frac{{{x}^{2}}}{{{A}^{2}}} \right){{\sin }^{2}}\,\,\phi \] \[\Rightarrow \frac{{{y}^{2}}}{{{B}^{2}}}+\frac{{{x}^{2}}}{{{A}^{2}}}\left( {{\cos }^{2}}\,\phi +{{\sin }^{2}}\,\,\phi \right)-\frac{2xy}{AB}\cos \,\,\phi ={{\sin }^{2}}\,\phi \]\[\Rightarrow \frac{{{x}^{2}}}{{{A}^{2}}}+\frac{{{y}^{2}}}{{{B}^{2}}}-\frac{2xy}{AB}\cos \,\phi ={{\sin }^{2}}\,\phi \] ?(3) When, \[\phi =\frac{\pi }{2},\,\,\cos \,\phi =0,\,\sin \,\phi =1\] Hence, Eq. (3) reduces to \[\frac{{{x}^{2}}}{{{A}^{2}}}+\frac{{{y}^{2}}}{{{B}^{2}}}=1\] This presents the equation of symmetrical ellipse.You need to login to perform this action.
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