Phase Difference Between Two Oscillators

Two oscillators of the same frequency can be in phase, out of phase, or anywhere in between. The phase difference $\varphi$ tells you how much one is "ahead of" the other in time. A phase difference of $\pi/2$ (90°) is particularly common — it's the offset between the position and velocity of a single oscillator.

The Math

If two oscillators have positions:

$x_1(t) = A\sin(\omega t) \qquad x_2(t) = A\sin(\omega t + \varphi)$

The phase difference is $\varphi$. When $\varphi = 0$, they move identically. When $\varphi = \pi$, they move oppositely (one at peak when the other is at trough). When $\varphi = \pi/2$, they're "quarter-cycle" offset — one is at the center when the other is at an extreme.

Step-by-Step Solution

Given: Two oscillators $x_1 = \sin(t)$ and $x_2 = \sin(t + \pi/2) = \cos(t)$.

Find: Tabulate their values at $t = 0,\, \pi/4,\, \pi/2,\, 3\pi/4,\, \pi$ and describe the relationship.

---

Step 1 — Tabulate the values.

| $t$ | $x_1 = \sin t$ | $x_2 = \cos t$ | |---|---|---| | $0$ | $0$ | $1$ (at peak) | | $\pi/4$ | $\sqrt{2}/2 \approx 0.707$ | $\sqrt{2}/2 \approx 0.707$ | | $\pi/2$ | $1$ (at peak) | $0$ | | $3\pi/4$ | $\sqrt{2}/2$ | $-\sqrt{2}/2$ | | $\pi$ | $0$ | $-1$ (at trough) |

(Tables aren't supported in the renderer; converted to a list above.)

• At $t = 0$: $x_1$ is at the center going up; $x_2$ is already at its peak.
• At $t = \pi/2$: $x_1$ has just reached the peak; $x_2$ has dropped back to the center.

$x_2$ is always one quarter cycle ahead of $x_1$. Equivalently, $x_2 = \sin(t + \pi/2) = \cos t$.

Step 2 — Identify the geometric pattern.

If you plot $(x_1, x_2)$ in 2D as $t$ varies, you trace out a circle. That's because:

$x_1^{2} + x_2^{2} = \sin^{2} t + \cos^{2} t = 1$

So a $\pi/2$ phase difference is the most "circular" pairing — it's how you get circular motion from two perpendicular SHM oscillators (this is called a Lissajous figure).

Step 3 — Other phase differences.

• $\varphi = 0$: $x_1 = x_2$ — both trace the same line in the $(x_1, x_2)$ plane.
• $\varphi = \pi$: $x_1 = -x_2$ — they trace the opposite line.
• $\varphi = \pi/2$: they trace a circle.
• Anything else: they trace an ellipse with axes tilted 45° from the coordinate axes.

These tilted ellipses are the famous Lissajous figures and were used by oscilloscope users in the analog era to measure phase relationships visually.

---

Answer: Two oscillators with a $\pi/2$ phase difference are a quarter cycle apart: one is at the peak when the other is at the center, and vice versa. Their parametric trajectory is a perfect circle in the $(x_1, x_2)$ plane:

$x_1^{2} + x_2^{2} = A^{2}$

For other phase differences, the trajectory is a Lissajous ellipse — a line for $\varphi = 0$ or $\pi$, a perfect circle for $\varphi = \pi/2$.

Try It

• Adjust the phase $\varphi$ from 0 to $2\pi$.
• The two waves shift relative to each other.
• The bottom panel plots the parametric trajectory $(x_1, x_2)$ — it morphs through ellipses, lines, and circles depending on $\varphi$.