Question

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.

• A. 300
• B. 400
• C. 600
• D. 900

Hint:

Use \(f_X/f_Y = (\text{vertical tangencies})/(\text{horizontal tangencies})\) for a Lissajous pattern in X-Y mode, then solve for \(f_Y\).

Answer 1

The Correct Option is D

Step 1: Recall the tangent-line rule for Lissajous patterns.
When two sinusoidal signals of different frequency are applied to the X and Y plates of an oscilloscope in X-Y mode, the closed trace touches a horizontal reference line, drawn tangent to its top or bottom, a certain number of times per cycle, and it touches a vertical reference line, tangent to its left or right edge, a certain number of times per cycle. The rule used to read frequency ratios off such patterns is:
\[ \frac{f_X}{f_Y} = \frac{\text{number of tangent points on a vertical line}}{\text{number of tangent points on a horizontal line}} \]
This works because the vertical extremes of the pattern, where the trace momentarily moves straight up or down and touches a vertical line, are set by the turning points of the X signal, while the horizontal extremes are set by the turning points of the Y signal. Counting these tangent points is the practical way an oscilloscope user measures an unknown frequency against a known reference frequency.

Step 2: Plug in the given tangency counts.
The problem states there are 3 horizontal tangencies and 2 vertical tangencies, so
\[ \frac{f_X}{f_Y} = \frac{2}{3} \]

Step 3: Solve for fY.
\[ f_Y = f_X \times \frac{3}{2} \]
With \(f_X = 600\) Hz:
\[ f_Y = 600 \times \frac{3}{2} = 900\ \text{Hz} \]

Step 4: Why the other options are wrong.
300 Hz (option A) and 400 Hz (option B) would follow only if the tangency counts were swapped or the ratio inverted the wrong way, for example using \(f_X/f_Y = 3/2\) instead of \(2/3\), or dividing instead of multiplying by \(3/2\). 600 Hz (option C) would be the answer only if the two frequencies were equal, which is not consistent with a pattern that shows different numbers of horizontal and vertical tangencies (3 versus 2); equal frequencies give a simple ellipse with equal tangency counts on both axes.

Final Answer:
The signal frequency on channel-Y is 900 Hz. \[ \boxed{f_Y = 900\ \text{Hz}} \]