A particle is executing two different simple harmonic motions, mutually perpendicular, of different amplitudes and having phase difference of $\frac{\mathrm{\pi }}{2}$ The path of the particle will be

• circular

• straight line

• parabolic

• elliptical

D.

elliptical

$$\begin{gathered} \mathrm{Let} \mathrm{two} \mathrm{perpendicular} \mathrm{simple} \mathrm{harmonic} \mathrm{motion}s \mathrm{are} \\ \mathrm{x} = \mathrm{A} \sin \left(\mathrm{\omega t}\right) ....\left(1\right) \\ \mathrm{y}= B \sin \left(\mathrm{\omega t} +\mathrm{\varphi }\right) ...\left(2\right) \\ \mathrm{from} \mathrm{equation} \left(1\right) \\ \sin \left(\mathrm{\omega t}\right) =\frac{\mathrm{x}}{\mathrm{A}} \\ \Rightarrow \sin { }^2\mathrm{\omega t} + {\cos }^2 \mathrm{\omega t} = 1 \\ \Rightarrow {\cos }^2 \mathrm{\omega t} = 1 - \frac{{\mathrm{x}}^2}{{\mathrm{A}}^2} \\ \Rightarrow \mathrm{cos\omega t} =\sqrt{1-\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}} \\ \left(2\right)\Rightarrow \frac{\mathrm{y}}{\mathrm{B}} = \sin \mathrm{\omega t} \mathrm{cos\varphi } + \mathrm{cos\omega t} \mathrm{sin\varphi } \\ \frac{\mathrm{y}}{\mathrm{B}}-\frac{\mathrm{x}}{\mathrm{A}}\mathrm{cos\varphi } =\sqrt{1-\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}} \mathrm{sin\varphi } \\ \mathrm{squaring} \mathrm{on} \mathrm{both} \mathrm{side} \\ \frac{{\mathrm{y}}^2}{{\mathrm{B}}^2} +\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}{\cos }^2\mathrm{\varphi } -2 \frac{\mathrm{y}}{\mathrm{B}} \frac{\mathrm{x}}{\mathrm{A}}\mathrm{cos\varphi } \\ =\left(1-\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}\right) {\sin }^2\mathrm{\varphi } \\ \frac{{\mathrm{y}}^2}{{\mathrm{B}}^2} +\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}\left({\cos }^2\mathrm{\varphi } +{\sin }^2\mathrm{\varphi }\right) \\ -\frac{2\mathrm{xY}}{\mathrm{AB}}\cos \mathrm{\varphi } = {\sin }^2\mathrm{\varphi } \\ \Rightarrow \frac{{\mathrm{y}}^2}{{\mathrm{B}}^2} +\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2} -\frac{2\mathrm{xy}}{\mathrm{AB}} \mathrm{cos\varphi }= {\sin }^2\mathrm{\varphi } \\ \mathrm{When} \mathrm{\varphi } =\frac{\mathrm{\pi }}{2}\Rightarrow \mathrm{cos\varphi } =0 \mathrm{sin\varphi } = 1 \\ \frac{{\mathrm{y}}^2}{{\mathrm{B}}^2}+\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2} = 1 ....\left(3\right) \end{gathered}$$