Linear Harmonic motion (simple oscillator) 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.

 A: 
$x(t)=A\sin(\omega t + \phi)$, 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:
$$
\phi = \tan^{-1}\left(\frac{\omega x(0)}{v(0)}\right)
$$
A: 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 $~v_0=0~$
$$\phi=\rm arctan2\left( {\frac {x_0\,\omega}{\sqrt {{x_0}^{2}{\omega}^{2}+{v_0}
^{2}}}},{\frac {v_0}{\sqrt {{x_0}^{2}{\omega}^{2}+{v_0}^{2}}}}
 \right) $$
