GATE 2026 IN Q.48: solution and answer

Official answer: DSource: Official answer key, GATE 2026 IN (IIT Guwahati)

1. In X-Y mode the spot is driven by $x=A\sin ({\omega }_{x}t)$ and $y=B\sin ({\omega }_{y}t+\varphi )$. Over one repeat period $T$ of the closed figure, the spot touches the top edge (a horizontal tangent line) once for every cycle of the Y signal. It touches the right edge (a vertical tangent line) once for every cycle of the X signal.

2. So the number of horizontal tangencies is $f_{y}T$ and the number of vertical tangencies is $f_{x}T$. Dividing one by the other gives the Lissajous rule $$\frac{f_y}{f_x}=\frac{\text{number of horizontal tangencies}}{\text{number of vertical tangencies}}$$

3. Tangencies are counted along one edge only. In a closed pattern the top and bottom edges always have the same count, and so do the left and right edges. So the 3 in the question is the count on one horizontal edge, and the 2 is the count on one vertical edge. (A total of 3 split across the top and bottom edges is impossible for a closed figure, because the two edges must carry equal counts.)

4. Substitute the given counts: $$\frac{f_y}{600\text{Hz}}=\frac{3}{2}$$

5. $$f_y=\frac{3}{2}\times 600\text{Hz}=900\text{Hz}$$

6. Check: $f_y:f_x=900:600=3:2$. The Y signal completes 3 cycles while X completes 2, so the figure touches the top edge 3 times and a side edge 2 times. That matches the question, so the answer is (D).