Phase Angle

In Simple Harmonic Motion (SHM), the phase angle is an important factor that shows where an oscillating object is at any given moment. It helps us understand how far along the oscillation is in its cycle. Usually measured in radians, the phase angle can vary from 0 to 2π radians and indicates how much the sinusoidal functions are offset. Grasping this concept is crucial for analyzing different types of oscillatory systems, whether they’re mechanical or electrical.

1.0Definition of Phase Angle

• The phase of a pulsating particle at any instant gives the state of the particles as considerations of its position and the direction of motion at that instant.
• It is equal to the argument of sine or cosine function occurring in the displacement equation of the S.H.M.Equation of S.H.M is represented as, $x=A\mathrm{Cos}\left(\omega t+{\varphi }_0\right)$

Phase of the particle is $\varphi =\omega t+{\varphi }_0$

• The phase $\varphi$ is a function of time t.It is usually expressed either as the fraction of the time period T or fraction of angle $2\pi$ that has elapsed since the vibrating particle last passed its mean position in the positive direction.
• Phase $\varphi$ gives an idea about the positions and the direction of motion of the pulsating particle.

2.0Initial Phase Or Epoch

The phase of a vibrating particle corresponding for time t=0.It describes the particle's initial state of motion while vibrating.

Example: Illustrate two circular motions, specifying the radius of each circle, the period of revolution, the initial position, and the direction of rotation. Derive the simple harmonic motion of the x-projection of the radius vector for the rotating particle P in both scenarios.

Solution:

(a) At t=0,OP makes an angle $45^{\circ }=\frac{\pi }{4}\mathrm{rad}$ with the positive direction of x-axis. After time t, it covers $\frac{2\pi }{T}t$ in the anticlockwise sense, and makes an angle of $\left(\frac{2\pi }{T}t+\frac{\pi }{4}\right)$ with the x-axis. The projection of OP on the x axis at time t is given by,

$x(t)=A\mathrm{Cos}\left(\frac{2\pi }{T}t+\frac{\pi }{4}\right)\text{ for }$

T=4 s, $x(t)=A\left(\frac{2\pi }{T}t+\frac{\pi }{4}\right)$

Which is a S.H.M of amplitude A,period 4s,and an initial phase $=\frac{\pi }{4}$

(b) In this case at t=0,OP makes an angle $90^{\circ }=\frac{\pi }{2}\mathrm{rad}$ with the x-axis. After a time t

It covers an angle of $\frac{2\pi }{T}t$ sense and makes an angle of $\frac{\pi }{2}-\frac{2\pi }{T}t$ with the x-axis.The projection of OP on the x-axis at time t is given by,

$x(t)=B\mathrm{Cos}\left(\frac{\pi }{2}-\frac{2\pi }{T}t\right)=B\mathrm{Sin}\left(\frac{2\pi }{T}t\right)$

For T=30 s, $x(t)=B\mathrm{Sin}\left(\frac{\pi }{15}-\frac{\pi }{2}\right)$

This represents a S.H.M of amplitude B,Period 30s,and an initial phase of $\frac{\pi }{2}.$

Dependence of Initial Angle

• Sinusoidal equation either sine or cosine equations
• Initial value of displacement (x) at t=0
• Initial directions of particle velocity

3.0Sample Questions On Phase Angle

Q-1.Define Phase Angle and its units?

Solution: The phase of a vibrating particle at any moment indicates its position and the direction of its motion at that instant.It is equal to the argument of sine or cosine function occurring in the displacement equation of the S.H.M.Equation, $x=A\mathrm{Cos}\left(\omega t+{\varphi }_0\right)$Phase of the particle is $\varphi =\omega t+{\varphi }_0$. It units is Radian

Q-2. If two S.H.M are represented by equations  $y_1=10\mathrm{Sin}\left[3\pi t+\frac{\pi }{4}\right]$ and $y_2=5[\mathrm{Sin}(3\pi t)+\sqrt{3}\mathrm{Cos}(3\pi t)]$ then find the ratio of their amplitudes and phase difference in between them.

Solution:

$y_2=5[\mathrm{Sin}(3\pi t)+\sqrt{3}\mathrm{Cos}(3\pi t)]$……..(1)

$5=A\mathrm{Cos}\varphi$

$5\sqrt{3}=A\mathrm{Sin}\varphi$

$A=\sqrt{5^2+(5\sqrt{3}{)}^2}=10$

$\mathrm{Tan}\varphi =\frac{5\sqrt{3}}{5}=\sqrt{3}\text{ so }\varphi =\frac{\pi }{3}$ 

Equation (1) becomes

$y_2=A\mathrm{Cos}\varphi \mathrm{Sin}(3\pi t)+A\mathrm{Sin}\varphi \mathrm{Cos}(3\pi t)$

$y_2=A\mathrm{Sin}(3\pi t+\varphi )$

$y_2=A\mathrm{Sin}\left(3\pi t+\frac{\pi }{3}\right)$

$\frac{A_1}{A_2}=\frac{10}{10}\Rightarrow A_1:A_2=1:1$

Phase difference = $\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}\mathrm{rad}$

 Q-3.A particle experiences two simple harmonic motions (S.H.M.) in the same direction, both with equal amplitudes and frequencies. If the resultant amplitude matches the amplitude of each individual motion, determine the phase difference between the two motions

Solution: Let amplitude of individual motion be A, and the resultant amplitude is A, if the phase difference between two motion is $\theta$,

$A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\mathrm{Cos}\theta }$

$A=\sqrt{A^2+A^2+2A\cdot A\mathrm{Cos}\theta }$

$A=\sqrt{2A^2+2A^2\mathrm{Cos}\theta }$

$\mathrm{Cos}\frac{\theta }{2}=\frac{1}{2}$

$\theta =\frac{2\pi }{3}$

Q-4. A particle performing SHM is found at its equilibrium position at t=1 sec and it is identified to have a speed of $0.25\mathrm{m}/\mathrm{s}$ at $t=2s$. If the period of oscillation is 8 s. Calculate phase angle.

Solution:

$x=A\mathrm{Sin}(\omega t+\varphi )$

At $t=1\mathrm{s}$ particle at mean position 

$0=A\mathrm{Sin}\left(\frac{2\pi }{8}\times 1+\varphi \right)\Rightarrow \varphi =-\frac{\pi }{4}$