Consider the equation for velocity of a body undergoing SHM-
$$ v(t)=-\omega A\sin(\omega t+\phi)$$
What does this negative sign mean?
Does it mean that velocity can be positive or negative at the same location in space?
OR,
Does it mean that velocity increases while displacement decreases in terms of their magnitude while their directions are opposite sometimes and sometimes remain the same? Does this actually have any significance?
Or, is it something else?
For a particle executing SHM, its displacement $x$ from its mean position at some time $t$ is given as $$ x(t) = A\cos (\omega t + \phi), $$ where $A$ is the amplitude, $\omega$ is the angular frequency and $\phi$ is the initial phase offset.
Its velocity is then given as (by differentiating the above with respect to time) $$ v(t) = -A\omega\sin (\omega t + \phi). $$
For simplicity, consider $\phi = 0$. This gives $x(t) = A\cos \omega t$ and $v(t)=-A\omega\sin \omega t$. So initially (at $t=0$), the particle is at $x=A$ (the extreme point in the positive $x$-direction), while the velocity is zero. This is expected since the velocity (of a particle executing SHM) at end points (or rather turning points) must be zero as the particle is changing directions. As time goes, the particle starts moving towards the origin (towards the left) and at $t = \pi/2\omega$, we find that $x=0$, i.e. the particle is at the origin. At this point of time, the velocity of the particle is $v = -A\omega$. The negative sign implies that the velocity is directed to the left and at the origin, it is maximum in magnitude. From now on until $t=\pi/\omega$, the displacement increases in the negative direction until $x=-A$ and the velocity (which is still in the negative $x$ direction) falls to zero. The particle then reverses its direction at $t>\frac{\pi}{\omega}$ and moves in the positive $x$-direction. In this way the particle keeps oscillating between the points $x=-A$ and $x=+A$.
The negative sign in the expression for velocity appeared because we differentiated the displacement (which had a cosine term) with respect to time, whose derivative is gave a minus sign. If we had started with the expression $$ x(t) = A\sin(\omega t + \phi), $$ the velocity would have been obtained as $$ v(t) = A\omega \cos (\omega t +\phi). $$ While there is no negative sign in this expression, that does not imply that the velocity is never negative. In both cases while the expressions for $x(t)$ and $v(t)$ vary, the physics is the same: the particle oscillates between two points and its velocity is directed such that it has to change sign at the turning points, otherwise the particle will continue its journey in one particular direction only and there will be no oscillatory motion.
