Our teacher said that displacement $x = A \sin(\omega t)$ in simple harmonic motion and velocity $v = dx/dt = A\omega \cos(\omega t)$ or $A\omega \sin(\omega t + \pi/2)$.

He also told us that $v$ leads $x$ by $\pi/2$, although one could see this by looking at the equation. I didn't quite understood what he meant by this.


2 Answers 2


They can be written as

$$x(t)= A\sin(\omega t)$$

$$v(t)= A\omega \cos(\omega t) = A\omega \sin(\omega t+\pi/2)$$

So when you make the graph of those two equations you'll see that your velocity leads the displacement with a phase $\pi/2$ since the prefactors only change the amplitude.


You can write down a simple code in Mathematica and obtain the following graph (for better understanding assume both prefactors are $1$, this way it will be easier for you to see.)

x-v graph

  • $\begingroup$ You can check the graph for a visual-aid. $\endgroup$
    – Monopole
    Feb 6, 2021 at 16:31

If you try to visualize the harmonic oscillator motion, you see that at maximum displacement (i.e. at the reversal point) the velocity is zero and vice versa. This means the velocity is out of phase by $\pi/2$.


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