Question

A particle undergoes mutually perpendicular simple harmonic motions given by $x=a\cos \omega t$ and $y=a\sin \omega t$. The trajectory of motion of the particle will be

• A. an ellipse
• B. a parabola
• C. a circle
• D. a straight line

Answer 1

The correct option is C a circle

Given the SHM equations in $x$ and $y$ directions are
$x=a\cos \omega t........\left(i\right)$
$y=a\sin \omega t........\left(ii\right)$
Squaring and adding equations $\left(i\right)$ and $\left(ii\right)$, we get
$x^2+y^2=a^2({\cos }^2\omega t+{\sin }^2\omega t)$
$\Rightarrow x^2+y^2=a^2$
$[\because {\cos }^2\omega t+{\sin }^2\omega t=1]$
This is the equation of a circle of radius $a$ with centre at the origin.
Hence, the trajectory of motion of the particle will be a circle.