# SHM phase relations between v,a,y

I was reading about the phase relation between the displacement , velocity and acceleration of a particle executing SHM.I understood that all three differ in phase by π/2 radian respectively . In my textbook it is written in addition to the above that

the velocity in SHM is in phase with acceleration , when particle is moving from extreme position to mean position and it is in opposite phase with acceleration when particle is moving from mean position to extreme position

How?

• I think what they mean to say is the domain where both are increasing is extreme to mean. And from mean to extreme velocity and acceleration are opposite (one increases and the other decreases). – Superfast Jellyfish Aug 26 at 8:17
• @SuperfastJellyfish but from extreme to mean also velocity increases till max. and acceleration decrease till min. =0 – user270071 Aug 26 at 9:01

The differential equation for SHM in 1D is given by $$\begin{equation} \ddot{y}+\omega^2y=0 \end{equation}$$
Its solution can be written as $$\begin{equation} y(t)=A\sin(\omega t+\phi) \end{equation}$$ where $$A$$ and $$\phi$$ is determined by the initial conditions (say $$\phi=0$$, which means that oscillator starts from equilibrium position). Then, $$\begin{equation} y(t)=A\sin(\omega t) \end{equation}$$
$$\begin{equation} v(t)=\dot{y}(t)=\omega A\cos(\omega t) \end{equation}$$
$$\begin{equation} a(t)=\dot{v}(t)=-\omega^2 A \sin(\omega t) \end{equation}$$
So the difference between $$v$$ and $$a$$ is always $$\pi/2$$ (rather $$3\pi/2$$). See the plot for $$\omega=2\pi$$ (black curve is $$v$$ and red curve is $$a$$), in going from equilibrium position to extreme, both $$v$$ and $$a$$ have opposite signs (blue region) and going from extreme to mean position they have the same sign (green region). Probably this is what you mean when you say 'phase'. 