Simple Harmonic Motion
5.0 Relation between Simple Harmonic Motion and Uniform Circular Motion
5.0 Relation between Simple Harmonic Motion and Uniform Circular Motion
If a body is performing uniform circular motion, then it must be doing periodic motion for all the points.
Consider a particle $P$ moving on a circle of radius $A$ with a constant angular velocity $\omega$. Let the center of the circle as the origin and two perpendicular diameters as the $X$ and $Y$ axes as shown in the Fig. SHM.3.
It is clear that as particle $P$ moves around the circle. The projection $Q$ oscillates between $S$ and $T$ or in other words when a particle $P$ moves with uniform circular motion, its projection on a diameter moves with SHM.
Consider a particle $P$ makes an angle of ‘$\phi$’ at $t=0$ with the $x$-axis.
Let at any time $t$ the particle $P$ makes an angle of ‘$\theta$’ with the $x$-axis as shown in the Fig SHM 4. Point $Q$ is the projection of point $P$ on the axis. $OP$ being the diameter ‘$A$’.
Further,
$$\begin{equation} \begin{aligned} OQ = OP\cos \theta \\ x = A\cos \left( {\omega t + \phi } \right) \\\end{aligned} \end{equation} $$
Note: $\theta = \left( {\omega t + \phi } \right)$ where $\phi$ is the angle particle makes at $t=0$ and ${\omega t}$ is the angle suspended by particle in time $t$.
The velocity of $P$ is perpendicular to $OP$ and has a magnitude of $v’ = ?A$. The component of $v’$ along $x$-axis is,
$$\begin{equation} \begin{aligned} v = - v'\sin \theta \\ v = - \omega A\sin \left( {\omega t + \phi } \right) \\\end{aligned} \end{equation} $$
Note: (-ve sign is used as velocity is along the negative $x$ axis)
The acceleration of $P$ is along the radius from $P$ to $O$ and has a magnitude, $a' = - {\omega ^2}A$. The component of $a’$ along the $x$-axis is $$\begin{equation} \begin{aligned} a = - a'\cos \theta \\ a = - {\omega ^2}A\cos \left( {\omega t + \phi } \right) \\\end{aligned} \end{equation} $$
Note: (-ve sign is used as acceleration is along the negative $x$ axis).
