Motion of a particle in x-y plane is described by a set of following equations $x = 4\sin \left( {{\pi \over 2} - \omega t} \right)\,m$ and $y = 4\sin (\omega t)\,m$. The path of the particle will be :
Solution
<p>$x = 4\sin \left( {{\pi \over 2} - \omega t} \right)$</p>
<p>$= 4\cos (\omega t)$</p>
<p>$y = 4\sin (\omega t)$</p>
<p>$\Rightarrow {x^2} + {y^2} = {4^2}$</p>
<p>$\Rightarrow$ The particle is moving in a circular motion with radius of 4 m.</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
This question is part of PrepWiser's free JEE Main question bank. 88 more solved questions on Oscillations are available — start with the harder ones if your accuracy is >70%.