Question

Prove that the motion of a particle along a circular path is simple harmonic and hence derive its displacement, velocity and acceleration.

Answer 1

$$\begin{gathered} Dearstudent \\ Anypointonacircleofradiusratanangle\theta withxaxisisgivenby \\ (x,y)\equiv (rcos\theta ,r\sin \theta ) \\ Letusconsider \\ y=r\sin \theta \\ y=r\sin \omega t \\ \frac{dy}{dt}=r\omega \cos \omega t \\ \frac{d^{2}y}{d{t}^2}=-r{\omega }^2\sin \omega t \\ \frac{d^{2}y}{d{t}^2}=-r{\omega }^{2}y \\ \frac{d^{2}y}{d{t}^2}=-ky \\ thisisthebasicequationofSHM \\ yistheydisplacementofparticlefromxaxis \\ y=r\sin \theta \\ \frac{dy}{dt}=r\omega \cos \omega t=velocity \\ \frac{d^{2}y}{d{t}^2}=-r{\omega }^2\sin \omega t=accelaration \end{gathered}$$