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The displacement of a particle performing simple harmonic motion is given by $x = A \sin(\omega t + \phi)$ , where $A$ is the amplitude, $\omega$ is the frequency, $t$ is the time, and $\phi$ is the phase constant. What is the significance of $\phi$. How is it used? Please explain the meaning of the phase constant

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  • $\begingroup$ The phase constant tells how much a signal is shifted along the x-axis. A phase constant of ϕ means that each value of the signal happens ϕ amount of time earlier. If the signal has a beginning, then a phase constant of ϕ means the signal occurs that much sooner. The phase constant is particularly significant when you have multiple signals, because having different phases can cause destructive interference. s1.thingpic.com/images/DF/QjQNCQ2eMfcLWJRa3k4gC6uB.png $\endgroup$
    – aebabis
    Feb 7, 2017 at 18:05

6 Answers 6

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The equation you state $$x=Asin(\omega t+\phi)$$ describes the displacement motion of a passive linear harmonic oscillator without loss. In other words there is no input or driving function. Whatever motion the oscillator exhibits is solely due to its initial conditions. $\phi$ in this case provides a point of reference in space for the oscillations.

But for the driven oscillator, $\phi$ provides a more significant role in terms of how efficiently energy is transferred from the driver to to the oscillator (system). If the driving force is in perfect phase with the system and pointing in the right direction, maximum energy is transferred at the harmonic resonant frequency. Either side of this point either leads or lags, decreasing the efficiency of energy transfer.

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  • $\begingroup$ That's an interesting point and you should elaborate. In particular for driven systems the two constant $A$ and $\phi$ are not determined by initial conditions but by the driving force. $\endgroup$
    – Borun Chowdhury
    Feb 7, 2017 at 16:56
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    $\begingroup$ We should be aware that the complete solution of a driven oscillator is the sum of the transient solution $A\exp(-\gamma t)\sin(\omega t+\phi)$ and the stationary solution $B\sin(\Omega t+\varphi)$. The constant $\varphi$ is the one which is related to the power of the oscillator. I usually call it phase lag, reserving the term phase constant to free parameter $\phi$. $\endgroup$
    – Diracology
    Feb 7, 2017 at 17:18
  • $\begingroup$ When you add a driving function things can begin to get very complex. Consider the simple pendulum. In terms of a real world system, a playground swing, the driving function is a pulse if it's a parent pushing the child. Much more complex when the child pumps with their legs. Each case, phase matters, determines increasing or decreasing amplitude $\endgroup$
    – docscience
    Feb 7, 2017 at 19:24
  • $\begingroup$ Not sure what you mean by "get very complex". Just take a harmonic oscillator in some state and then use the retarded Green's function to convolute with the applied force. $\endgroup$
    – Borun Chowdhury
    Feb 8, 2017 at 10:39
  • $\begingroup$ @BorunChowdhury Hah! things seem very simple when you ONLY look at the mathematics, the models. But if you look CLOSER at the physical reality, the real physics that are taking place, the means by which forces couple into the systems is quite complex. Even a father pushing his daughter on the playground swing! $\endgroup$
    – docscience
    Feb 8, 2017 at 18:03
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All the phase angle does is to give you a facility to decide on the displacement of the particle undergoing shm at time $t=0$ or any other time.

With your phase angle of $\phi$, assuming it to be positive, the graphs of $x_1 = a \sin (\omega t)$ (grey) and $x_2= A \sin (\omega t + \phi)$ (red) are shown below.

enter image description here

In this case the motion $x_2$ is in advance of the motion of $x_1$ by a time $t$ (shown in the diagram) or a phase angle of $\phi= \dfrac t T 2 \pi$ where $T$ is the period of the motion and equal to $\dfrac {2 \pi}{\omega}$.

So everything that particle with displacement $x_2$ does the particle with displacement $x_1$ does a time $t$ later.

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The equation of motion for a simple harmonic oscillator is

$$ \ddot x+\omega^2 x=0 $$

and the most general solution to this is

$$ x(t) = A_1 \cos \omega t + A_2 \sin \omega t $$

Note there are two constants of integration that correspond to the equation being a second order differential equation. More physically, the velocity is given by

$$ v(t) = - \omega A_1 \sin \omega t + \omega A_2 \cos \omega t $$

and the two constants of integration are fixed by the configuration of the system at any given time. Said differently, if you know the position and velocity at time $t_0$ you can solve for $A_1$ and $A_2$. You should do that as an exercise.

Now note that the expression you have can be written as

$$ x(t) = A \sin \phi \cos \omega t + A \cos \phi \sin \omega t $$

and you can therefore relate $A$ and $\phi$ to $A_1$ and $A_2$ and from there to the position and velocity at time $t_0$.

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  • $\begingroup$ So does it represent the amplitude positions of the particle. ? Also, how do I determine the value of $\phi$ $\endgroup$
    – user140805
    Feb 7, 2017 at 16:19
  • $\begingroup$ All the answers you need are in the post above. If you know the "initial conditions" at a certain time or "boundary conditions" at two different times you can find $A$ and $\phi$. However, the way you phrased the question it sounded like an attempt to understand a concept. Your comment makes it look like a homework problem. $\endgroup$
    – Borun Chowdhury
    Feb 7, 2017 at 16:35
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In the basic SHM equation, you get x=Asin(ωt) where at t=0, the object is at mean position or zero displacement. Now, what is the significance of the angle inside the sine function? It gives you the position of the particle performing SHM. When the angle is π/2, the displacement is maximum i.e A. When it is π, the displacement is once again 0. So, for the equation Asin(ωt+ϕ), it simply means that the SHM does not begin at x=0 and the position at t=0 is Asin(ϕ) (depending upon the value of ϕ it could be A,A/2 anything).

If the initial position is S, then ϕ=sin^-1 (S/A)

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    $\begingroup$ This answer in incomplete as you get the same position at $t=0$ for $\phi$ and $\pi- \phi$. You need to also know the initial velocity. Said differently the initial velocity is $A\omega \cos \phi$. $\endgroup$
    – Borun Chowdhury
    Feb 7, 2017 at 16:42
  • $\begingroup$ Yeah, of course. I just wanted to give a better idea about the significance of ϕ $\endgroup$
    – TheFool
    Feb 7, 2017 at 16:47
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What is the significance of $\phi$?

The phase angle $\phi$ represents the relation between the displacement and velocity of the simple harmonic oscillator at the point in time arbitrarily designated as $t=0$. In particular,$$\tan\phi = \omega \frac{x(0)}{v(0)}$$ The point in time at which $t$ is zero is completely arbitrary. With a different time axis given by $t' = t-t_0$, the state of the SHO can be expressed as $x(t) = A \sin(\omega t' + \phi')$, where $\phi' = \phi + \omega t_0$.

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Based on a point raised by @docscience this answer addresses the phase in terms of "initial conditions" introduced by driving forces. In fact one can think of this as answering how the SHO was set in motion in the first place.

The position of a simple harmonic oscillator at time $t$ that experienced force at time $t'$ and that was at rest in the far past

$$ \lim_{t\to -\infty} x(t)=0 \\ \lim_{t \to -\infty} \dot x(t)=0 $$

is given by

$$ x(t) = \int_{-\infty}^t \frac{1}{\omega} \sin (\omega (t-t')) f(t') $$

This has been obtained by using retarded Green's function for the SHO details of which can be found elsewhere but one can check that this satisfies the SHO equation of motion.

(1) For the simplest case lets take the case of a pulse of force at time $t'=t_0$ then we get

$$ x(t) = \frac{1}{\omega} \sin(\omega( t- t_0)) \Theta(t-t_0) $$ where $\Theta(t-t_0)$ is the Heavyside step function. Thus we see that the oscillator is at rest for $t<t_0$ and after that the 'phase' is $-\omega t_0$.

(2) Now lets take the case of two pulses at times $t_0$ and $t_1$ with amplitude $f_0$ and $f_1$ i.e.

$$ f(t)=f_0 \delta(t-t_0) + f_1 \delta(t-t_1) $$

with $t_1>t_0$. Its easy to see the solution is

$$ x(t)=\frac{f_0}{\omega} \sin(\omega( t- t_0)) \Theta(t-t_0) + \frac{f_1}{\omega} \sin(\omega( t- t_1)) \Theta(t-t_1) $$

Here is where we see the meaning of phase clearly: If we take $f_1 = f_2$ then we see it is possible to choose $t_0$ and $t_1$ such that the two pulses are "in-phase" and the amplitude doubles or "out-of-phase" such that the amplitude cancels and the second pulse just stops the SHO. These correspond to $\omega(t_1- t_0)=2 n \pi$ and $\omega(t_1-t_0)=n\pi$ for $n$ an odd integer.

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