A closed Lissajous pattern observed on an oscilloscope in the \(X-Y\) mode has three horizontal tangencies and two vertical tangencies in steady state. If the signal frequency on channel-\(X\) is 600 Hz, then the signal frequency on channel-\(Y\) is __________ Hz.
Show Hint
Use \(f_X/f_Y = (\text{vertical tangencies})/(\text{horizontal tangencies})\) for a Lissajous pattern in X-Y mode, then solve for \(f_Y\).
Cross-checking the tangency rule with a known example, then applying it. Why the rule works.
Picture the X-signal as $x(t)=\sin(2\pi f_X t)$ and the Y-signal as $y(t) = \sin(2\pi f_Y t + \phi)$. The trace touches a vertical tangent line whenever $x(t)$ reaches a local maximum or minimum, a turning point of the X-signal, and over one full repeat of the pattern this happens a number of times proportional to $f_X$. Likewise, the trace touches a horizontal tangent line a number of times proportional to $f_Y$. Since the counting is normally done on ONE tangent line, say the rightmost vertical line or the topmost horizontal line, the two counts end up in the ratio
\[
(\text{vertical tangencies}) : (\text{horizontal tangencies}) = f_X : f_Y
\]
Check this on a simple case: if $f_Y = 2f_X$, the pattern becomes a sideways figure-eight that touches a single vertical line once but touches a single horizontal line twice. That gives (vertical : horizontal) $= 1:2 = f_X:f_Y$, confirming the rule above.
Applying it to this problem.
Here horizontal tangencies $=3$ and vertical tangencies $=2$, so
\[
2 : 3 = f_X : f_Y
\]
\[
\frac{f_Y}{f_X} = \frac{3}{2}
\]
\[
f_Y = \frac{3}{2}\times 600 = 900 \text{ Hz}
\]
Sanity check.
Since $f_Y > f_X$ (900 > 600), the Y trace completes more oscillations than the X trace in one repeat of the pattern, matching the fact that the pattern shows MORE horizontal tangencies (3, tied to the faster Y-signal's turning points) than vertical tangencies (2, tied to the slower X-signal). This directional check rules out the smaller options 300 Hz and 400 Hz, which would both require $f_Y < f_X$.
\[ \boxed{f_Y = 900 \text{ Hz}} \]
