Question

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

• A. Circular
• B. Straight line
• C. Parabolic
• D. Elliptical

Answer 1

Answer: (D)

Let two perpendicular simple harmonic motions are
$x=A\sin (\omega t)...(1)$
$y=B\sin (\omega t+\varphi )...(2)$
From Eq. (1) $\sin (\omega t)=\frac{x}{A}$
$\Rightarrow \cos \omega t=\sqrt{1-\frac{x^2}{A^2}}$
From Eq. (2)
$\frac{y}{B}=\sin (\omega t)\cos \varphi +\cos (\omega t)\sin \varphi$
$=\left(\frac{x}{A}\right)\cos \varphi +\sqrt{1-\frac{x^2}{A^2}}\sin \varphi$
$\Rightarrow \frac{y}{B}-\frac{x}{A}\cos \varphi =\sqrt{1-\frac{x^2}{A^2}}\sin \varphi$
On squaring, we get
$=\frac{y^2}{B^2}+\frac{x^2}{A^2}{\cos }^2\varphi -\frac{2xy}{AB}\cos \varphi =\left(1-\frac{x^2}{A^2}\right){\sin }^2\varphi$
$\Rightarrow \frac{y^2}{B^2}+\frac{x^2}{A^2}\left({\cos }^2\varphi +{\sin }^2\varphi \right)-\frac{2xy}{AB}\cos \varphi ={\sin }^2\varphi$
$\Rightarrow \frac{x^2}{A^2}+\frac{y^2}{B^2}-\frac{2xy}{AB}\cos \varphi ={\sin }^2\varphi ...(3)$
When, $\varphi =\frac{\pi }{2},\cos \varphi =0,\sin \varphi =1$
Hence, Eq. (3) reduces to
$\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$
This presents the equation of symmetrical ellipse.