Simple Harmonic Motion
7.0 How to Write the Simple Harmonic Motion Equation
7.0 How to Write the Simple Harmonic Motion Equation
This type of questions are solved by using relation between SHM and uniform circular motion.
Consider the center of the circle as $O$ and the perpendicular diameters as $X$ and $Y$ axes.
Let ‘displacement ($x$)’ be along $Y$ axes and ‘velocity ($v$)’ be along $X$ axes.
So we can assume the sign convention of displacement and velocity in all the four quadrants of a circle as shown in the fig SHM 10.
Let us understand the above two topics with the help of some examples.
Example 1. With reference to the position of the particle from Fig (a) at time $t=0$. Write the equation of SHM at any time $t$.
Solution: Since, the displacement of the particle is on the right side of the mean position, therefore it is $+ve$, and the direction of velocity is towards right, so it is $+ve$.
Taking ‘displacement’ along $Y$ axes and ‘velocity’ along $X$ axes, for displacement and velocity to be $+ve$, it lie in ‘first quadrant’.
With reference to the fig. (b).
$$\begin{equation} \begin{aligned} \sin \theta = \frac{a}{{2a}} = \frac{1}{2} \\ \theta = \frac{\pi }{6}\;({\text{at time }}t = 0) \\\end{aligned} \end{equation} $$
So, the equation of SHM at any time ‘$t$’ is, $$x = 2a\sin \left( {\omega t + \frac{\pi }{6}} \right)$$
Example 2.
Solution: Since displacement of the particle is on the right side of the mean position, therefore the displacement is $+ve$ and the direction of velocity is towards left, so it is $–ve$.
Taking ‘displacement’ along $Y$ axes and ‘velocity’ along $X$ axes, for displacement to be $+ve$ and velocity to be $-ve$, it lie in ‘second quadrant’.
With reference to the fig. (b),
In triangle $PQO$, angle $POQ$ is $\phi$,
$$\begin{equation} \begin{aligned} \sin \phi = \frac{a}{{2a}} = \frac{1}{2} \\ \phi = \frac{\pi }{6}\;({\text{at time }}t = 0) \\\end{aligned} \end{equation} $$$$\therefore \theta = \frac{\pi }{3} + \frac{\pi }{2} = \frac{{5\pi }}{6}$$
So, the equation of SHM at any time ‘$t$’ is,
$$x = 2a\sin \left( {\omega t + \frac{{5\pi }}{6}} \right)$$
