Question

Two simple harmonic motion of same amplitude, same frequency and a phase difference of $\pi$ are superimposed at right angles to each other on a particle, the particle will describe a

• A. circle
• B. ellipse
• C. figure of eight
• D. straight line

Answer 1

Answer: (B)

ellipse

Here, $x=a\sin \omega t$ and $\cos \omega t=\sqrt{1-\frac{x^2}{a^2}}$
$y=b\sin (\omega t+\varphi )=b\sin \omega t\cos \varphi +b\cos \omega t\sin \varphi =b\frac{x}{a}\cos \varphi +b\sqrt{1-\frac{x^2}{a^2}}\sin \varphi$
or $(\frac{y}{b}-\frac{x}{a}\cos \varphi )=\sqrt{1-\frac{x^2}{a^2}}\sin \varphi$
squaring both sides we get, $\frac{y^2}{b^2}+\frac{x^2}{a^2}-\frac{2xy}{ab}\cos \varphi ={\sin }^2\varphi$
given, $a=b$ and $\varphi =\frac{\pi }{4}$
so, $\frac{y^2}{a^2}+\frac{x^2}{a^2}-\frac{2xy}{a^2}.\frac{1}{\sqrt{2}}=\frac{1}{2}$
This is the equation of an ellipse.