Phase in S.H.M.

Oscillations · Maharashtra HSC

Theory

What phase means

Phase in S.H.M. (or for any motion) is basically the state of oscillation. To know the state of oscillation we need to know three things at that instant of time:

the displacement (position),
the direction of velocity, and
the oscillation number (during which oscillation it is happening).

Why displacement or velocity alone will not do

Knowing only the displacement is not enough, because at a given position there are two possible directions of velocity (except at the extreme positions), and the same position repeats in successive oscillations.
Knowing only the velocity is not enough, because there are two different positions for the same velocity (except at the mean position).
Both of these repeat for successive oscillations.

Hence, to know the phase, we need a quantity that is continuously changing with time. All the quantities of a linear S.H.M. ( $x$ , $v$ , $a$ ) are projections, on a diameter, of the corresponding quantities of the reference circular motion. The angular displacement of that reference motion can therefore be used as the phase of the S.H.M., since it varies continuously with time. In this case it is called the phase angle: $\theta=\omega t+\phi$

Epoch

The constant $\phi$ is the phase at $t=0$ . The physical quantity which describes the state of oscillation of a particle performing S.H.M. at the start of the motion is called the epoch (or initial phase) of the S.H.M.

Special cases

Phase $\theta$	State of the particle
$0$	At the mean position, moving to the positive, at the beginning of the first oscillation
$2\pi^c$ or $360^\circ$	The same state at the beginning of the second oscillation, and so on
$\pi^c$ or $180^\circ$	During the first oscillation, at the mean position and moving to the negative
$(2\pi+\pi)^c$ or $(360+180)^\circ$	The same state during the second oscillation
$\dfrac{\pi^c}{2}$ or $90^\circ$	At the positive extreme position during the first oscillation
$\left(2\pi+\dfrac{\pi}{2}\right)^c$ or $(360+90)^\circ$	At the positive extreme during the second oscillation
$\dfrac{3\pi^c}{2}$ or $270^\circ$	At the negative extreme position during the first oscillation
$\left(2\pi+\dfrac{3\pi}{2}\right)^c$ or $(360+270)^\circ$	At the negative extreme during the second oscillation

Working with a phase angle

Reading a state from a phase — split off whole multiples of $360^\circ$ first. For a phase angle of $1110^\circ$ : $1110^\circ=3\times360^\circ+30^\circ$ Three complete revolutions plus something indicates the 4th oscillation, and $x=A\sin30^\circ=\dfrac{A}{2}$ . So the particle is at $+\dfrac{A}{2}$ during its 4th oscillation, moving towards the positive extreme.

Finding a phase from a state — take the inverse sine, then place it in the correct quadrant, then add $2\pi$ for each completed oscillation. While completing its third oscillation a particle is at $-\dfrac{\sqrt3}{2}A$ , heading to the mean position: $\sin^{-1}\left(-\dfrac{\sqrt3}{2}\right)\Rightarrow\text{reference angle }\dfrac{\pi}{3}$ Coming from the negative side towards the mean position places it in the fourth quadrant of that oscillation, giving $-\dfrac{\pi^c}{3}$ . As it is the third oscillation, two complete oscillations are over: $\theta=2\times2\pi-\dfrac{\pi}{3}=4\pi-\dfrac{\pi}{3}=\dfrac{11\pi^c}{3}$

Standard results

Phase angle: $\theta=\omega t+\phi$ , in radian
Epoch $\phi$ is the phase at $t=0$
$\theta=2\pi N+\theta_0$ : $N$ complete oscillations are over, the particle is in oscillation number $(N+1)$
$\theta=0$ : mean position moving positive; $\dfrac{\pi}{2}$ : positive extreme; $\pi$ : mean position moving negative; $\dfrac{3\pi}{2}$ : negative extreme