For two sine waves of different frequencies, how does their phase difference behave over time if both continue at their respective constant frequencies?

Introduction:
Phase describes the relative timing of periodic signals. When two sinusoids have different frequencies, their relative alignment does not remain fixed. This question tests conceptual understanding of phase evolution between unequal-frequency sinusoids.

Given Data / Assumptions:

• Two ideal sine waves in steady state.
• Frequencies f1 and f2 are constant but not equal.
• No noise or modulation; purely deterministic signals.

Concept / Approach:
Let phases be φ1(t) = 2 * pi * f1 * t + φ1,0 and φ2(t) = 2 * pi * f2 * t + φ2,0. The phase difference Δφ(t) = φ1(t) − φ2(t) = 2 * pi * (f1 − f2) * t + (φ1,0 − φ2,0). If f1 ≠ f2, the term 2 * pi * (f1 − f2) * t grows linearly with time, making Δφ(t) change continuously.

Step-by-Step Solution:
1) Write each sinusoid's instantaneous phase as 2 * pi * f * t plus a constant.2) Subtract to get the relative phase Δφ(t).3) Recognize the linear-in-time term proportional to frequency difference.4) Conclude: unequal frequencies imply a continuously changing phase difference.

Verification / Alternative check:
Oscilloscope Lissajous figures drift when frequencies are unequal, visibly demonstrating the evolving phase relationship as the figure slowly rotates or morphs over time.

Why Other Options Are Wrong:
Equal to their frequency differences: confuses rate of phase change (proportional to frequency difference) with phase itself.Difference in fixed time displacement: applies only when frequencies are the same.The same throughout time: true only for equal-frequency (and stable) sinusoids.

Common Pitfalls:
Mixing phase (an angle that evolves) with frequency (a rate). A small frequency mismatch causes steady, unbounded drift of phase difference.

Final Answer:
constantly changing