Phase Angle In SHM

Question

In a simple harmonic motion a trial function $A\sin(\omega t+\varphi)$ is used where $\varphi$ denotes phase angle. What is the work of $\varphi$ in the equation. Doesn't it work for $A\sin(\omega t)$ only?

Answer 1

Phase angle as the name suggests simply describes the initial condition of the simple harmonic motion. If you begin the motion at any point other than the zero or equilibrium position then you need a phase shift angle as when $t=0$, $A\sin(\omega t) = A\sin(\omega ( 0)) = A\sin(0) = 0$.

Thus in your system if the output is anything but 0 at $t=0$, then you need a phase shift angle

This works for any trigonometric function. If the output is different than the trig function would suggest at $t=0$, then a phase shift angle is required. All in all, that value simply describes how much the oscillation or period is "offset" from the regular trigonometric function.

Answer 2

The general solution of $\ddot{x}=-\omega^2 x$ can be written in several ways, as linear combinations of $\exp\pm i\omega t$, or as linear combinations of $\cos\omega t,\,\sin\omega t$, or (rewriting the latter with compound angle formulae) as $A\sin(\omega t+\varphi)$. Indeed, the last of these follows from rewriting a linear combination of $\exp\pm i\omega t$ as one of $\exp\pm i(\omega t+\varphi)$, as this just rescales the original basis elements by factors of $\exp\pm i\varphi$.

Answer 3

A point rotates on a circle radius A centered at origin starting from x-axis. Its projection on x-axis is SHM.

It has a differential equation and solution

$$ \ddot x+ \omega^2 x =0 ,\, x=A \cos (\omega t )$$

However we can start from any point on the circle when its radius vector makes $\varphi$ to x- axis and still it is an SHM, having the same differential equation but the shift is shown as

$$ x=A \cos (\omega t + \varphi )$$

This altered start motion with an offset obeys the same DE. The angle $\varphi$ is referrred to as PhaseShift