We know that for a simple harmonic linear oscillator, the displacement is given by $x(t)=A\sin(\omega t + \phi)$, where $\phi$ denotes the phase angle. Now as per my understanding this $\phi$ is only significant when considering SHM in form of a sinusoidal wave. Is there any physical meaning in reality. Is there a way to measure the phase angle in reality just by virtue of the particle's (which is oscillating) position with respect to the mean position?

Some Assumptions

  • I am considering SHM in only one axis.
  • I am also considering SHM linearly for ex Spring Block System.
  • $\begingroup$ Yes, this is important in adding waves. $\endgroup$ Jan 26 at 4:22
  • $\begingroup$ Yes,I am aware of this but as you might know that for in circular motion the phase angle is analogously the angular displacement which a particle performs with respect to the center. Is there a similar thing with linear oscillator? $\endgroup$ Jan 26 at 4:51
  • $\begingroup$ You can get the phase by measuring the velocity and displacement over some given time period and plotting them. $\endgroup$
    – Triatticus
    Jan 26 at 5:18
  • $\begingroup$ Yes you can assume the position of an object doing SHM by just observing the phase. $\endgroup$ Jan 26 at 5:33

2 Answers 2


$x(t)=A\sin(\omega t + \phi)$, where $\phi$ denotes the phase angle... Is there any physical meaning in reality... Is there a way to measure the phase angle in reality just by virtue of the particle's (which is oscillating) position with respect to the mean position?

The mean position of what you wrote above is zero.

The phase angle $\phi$ (along with the other parameters $A$ and $\omega$) tells you the initial position and velocity.

You can re-write $x(t)$ as $$ x(t) = A\left(\sin(\omega t)\cos(\phi)+\cos(\omega t)\sin(\phi)\right)\;, $$ to see that the initial position is given by $$ x(0) = A\sin(\phi)\;, $$ and the initial velocity is given by $$ v(0) = A\omega\cos(\phi)\;. $$

You can solve for $\phi$ like: $$ \phi = \tan^{-1}\left(\frac{\omega x(0)}{v(0)}\right) $$

  • $\begingroup$ If we take a spring block system for example. With what respect are we measuring this phase angle? $\endgroup$ Jan 26 at 5:41
  • $\begingroup$ @TheCuriousOne With respect to mean position on axis of motion, when observation started. $\endgroup$ Jan 26 at 6:54
  • $\begingroup$ But I just don’t get it the angle only remains 0 radians the whole time where is the angle formed? $\endgroup$ Jan 26 at 7:52
  • $\begingroup$ @TheCuriousOne When you projected radius on any one axis, you get similar result of mass-spring. The phase constant is when one start observation and radius is not aligned to fixed axis along which projection is measured as displacement of oscillator. Linear doesn't mean moving in straighr line but whose responses or stimuli can be added linearly, as harmonics in fourier series. $\endgroup$ Jan 26 at 9:31
  • $\begingroup$ So what you are trying to say is that we are considering a point(imaginary) through which when we connect with the mean position we get angle? $\endgroup$ Jan 26 at 10:41

SHM has this differential equation

$$\ddot x(t)+\omega^2\,x(t)=0\tag 1$$

to solve this differential equation you need two initial conditions.

$$x(0)=x_0~,\dot x(0)=v_0$$

with your Ansatz $~x(t)=A\,\sin(\omega\,t+\phi)~$ in equation (1) you obtain $~0=~0~$ thus $~x(t)~$ is the solution of the SHM. to obtain the constants $~A~,\phi~$ you have two equations

$$x(t=0)=A\,\sin(\phi)=x_0\\ \dot x(t=0)=A\,\omega\cos(\phi)=v_0~,$$

for the unknowns $~A~,\phi~\Rightarrow$

$$A=\frac 1\omega\,\sqrt{\omega^2\,x_0^2+v_0^2}\\ \phi=\arctan\left(\omega\frac{x_0}{v_0}\right)$$

this work also for $~v_0=0~$

$$\phi=\rm arctan2\left( {\frac {x_0\,\omega}{\sqrt {{x_0}^{2}{\omega}^{2}+{v_0} ^{2}}}},{\frac {v_0}{\sqrt {{x_0}^{2}{\omega}^{2}+{v_0}^{2}}}} \right) $$


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