What is the significance of the phase constant in the Simple Harmonic Motion equation? 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
 A: 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.
A: 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.

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.
A: 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 $t_0$.
A: 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)
A: 
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 $\phi' = \phi + \omega t_0$.
A: 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.
