2

I came across a paragraph that, as I interpreted, conveyed that:

1). The phase of a vibrating particle is the ratio of the displacement of the vibrating particle to its amplitude, at any instant.

2). It is the fraction of the time interval that has elapsed after the particle crossed the mean position of rest, whilst travelling in the positive Y direction (USING X and Y axis).

3). Refers to phase angle.

I have difficulty understanding and accepting points 1 and 2. I normally hear of phase dealing with the phase angle...the first two points are new to me. Could someone please explain them to me? I think point 2 has to do with the time period that the particle takes to vibrate simple harmonically.

rert588
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2 Answers2

1

Phase means phase angle, so interpretation (3) is correct. When you describe the oscillation in the form $y=A\sin(\omega t - kx)$ then the argument $\omega t-kx=\phi$ is the phase.

Interpretation (1) is referring to the fraction $\frac{y}{A}=\sin(\omega t-kx)$, so this is not the phase.

Interpretation (2) is partially correct. If you are looking at a wave at a fixed position, say $x=0$, then the phase angle is $\phi=\omega t=2\pi \frac{t}{T}$ where $T$ is the period. So phase is proportional to time, and is a fraction of the period.

sammy gerbil
  • 27,277
1

Phase is a term used when there is a particle which is vibrating with a period $T$ to describe what fraction of a complete oscillation the particle has completed when it is at some new position.

So the clock is started when the particle is at a particular point in the oscillation and moving in a particular direction.
The time taken for the particle to reach that point again whilst moving in the same direction is called the period $T$.
If the particle is in some new position after a time $t$ the en fraction of an oscillation that the particle has undergone is $\frac t T$.

In your statement $(2)$ the clock is started when the particle is at position $y=0$ and moving in the positive y-direction and the time $t$ is measured when the particle is at some new position.
The fraction of a complete oscillation that the particle has undergone when in its new position is $\frac t T$ and this is called the phase.

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If the particle was at position $y=0$ and moving in the opposite direction to that when the clock was started the time taken would be $\frac T 2$ and the fraction of a complete oscillation that the particle has undertaken in that time is the pahse $= \frac{T/2}{2}= \frac 12$.

Often it is more convenient not to talk about fractions of a complete oscillation but to define one complete oscillation as either $360^\circ$ or $2 \pi^{\rm c}$ and then the phase angle is defined as $\frac t T \times 360^\circ$ or $\frac t T \times 2 \pi ^{\rm c}$.

For the motion depicted below using your statement $(2)$ the fraction of a complete oscillation that the particle has undertaken between time $1_1$ and time $t_2$ is $\frac {t_2-t_1}{T}$.
Note that this is the same fraction of a complete oscillation that the particle has undergone between time $t_3$ and time $t_4$ because $t_4-t_3=t_2-t_1$

Your statement $(1)$ is not correct because the speed of the oscillating particle is not constant so the position of the particle is not proportional to time.

Suppose that $t_2 - t_1 = \frac T 8$ and the graph shown in the diagram is sinusoidal.
This means that in that period of time the particle has undergone $\frac 1 8$th of a complete oscillation, phase $= \frac 1 8$.
If the amplitude of motion of the particle is $A$ then at time $\frac T 8$ the position of the particle is $y = \frac {A}{\sqrt 2}$ then $\frac {A/\sqrt 2}{A}= \frac {1}{\sqrt 2}$ is not the fraction of a complete oscillation that the particle has completed in a time $\frac T 8$.

Farcher
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