Two simple harmonic motions, as shown, are at right angles. They are combined to form Lissajous figures. $$x(t) = A\sin(at + \delta)$$, $$y(t) = B\sin(bt)$$. Identify the correct match below.

We are given two perpendicular simple harmonic motions,

$$x(t)=A\sin\left(at+\delta\right),\qquad y(t)=B\sin\left(bt\right).$$

To know the Lissajous curve we eliminate the time variable. The most convenient way is to compare the two arguments. We therefore begin by considering the special situation when the two angular frequencies are the same, that is, when $$a=b.$$ From the four alternatives, Options B, C and D have $$a=b,$$ whereas Option A has $$a=2b.$$ We treat the equal-frequency case first and afterwards check each option.

Put $$a=b$$ and introduce a simpler symbol $$\alpha$$ by writing

$$\alpha = at \;\;(\text{because } a=b).$$

Hence

$$x=A\sin\left(\alpha+\delta\right),\qquad y=B\sin\alpha.$$ We now expand the shifted sine with the trigonometric angle-addition formula

$$\sin(\alpha+\delta)=\sin\alpha\cos\delta+\cos\alpha\sin\delta.$$

Substituting this expansion into the expression for $$x$$, we obtain

$$x=A\left(\sin\alpha\cos\delta+\cos\alpha\sin\delta\right).$$

Now divide both $$x$$ and $$y$$ by their respective amplitudes so that the expressions become dimensionless:

$$\frac{x}{A}=\sin\alpha\cos\delta+\cos\alpha\sin\delta,\qquad \frac{y}{B}=\sin\alpha.$$

We wish to eliminate both $$\sin\alpha$$ and $$\cos\alpha$$. Start by writing $$\sin\alpha$$ directly from the second equation:

$$\sin\alpha=\frac{y}{B}.$$

Next solve the first equation for $$\cos\alpha$$:

$$\frac{x}{A}-\sin\alpha\cos\delta=\cos\alpha\sin\delta,$$ so $$\cos\alpha=\frac{1}{\sin\delta}\left(\frac{x}{A}-\frac{y}{B}\cos\delta\right).$$

We now use the Pythagorean identity

$$\sin^{2}\alpha+\cos^{2}\alpha=1.$$

Substituting the two expressions we just derived gives

$$\left(\frac{y}{B}\right)^{2} +\left[\frac{1}{\sin\delta}\left(\frac{x}{A}-\frac{y}{B}\cos\delta\right)\right]^{2}=1.$$

Multiply by $$\sin^{2}\delta$$ to clear the denominator:

$$\sin^{2}\delta\left(\frac{y}{B}\right)^{2} +\left(\frac{x}{A}-\frac{y}{B}\cos\delta\right)^{2}=\sin^{2}\delta.$$

Finally expand the square on the right and gather terms in a neat symmetrical form:

$$\left(\frac{x}{A}\right)^{2} -2\frac{x}{A}\frac{y}{B}\cos\delta +\left(\frac{y}{B}\right)^{2}=1.$$

This is the general Cartesian equation of an ellipse whose precise shape depends on $$A,\,B$$ and $$\delta$$. Special choices of these parameters reduce the ellipse to other well-known curves, which we now list:

• If $$A=B$$ and $$\delta=\frac{\pi}{2},$$ the mixed term $$-2(x/A)(y/B)\cos\delta$$ vanishes because $$\cos\left(\frac{\pi}{2}\right)=0$$, and the equation reduces to

$$\left(\frac{x}{A}\right)^{2}+\left(\frac{y}{A}\right)^{2}=1,$$ which is a circle.

• If $$\delta=0,$$ the equation becomes

$$\left(\frac{x}{A}-\frac{y}{B}\right)^{2}=0 \;\Longrightarrow\; \frac{x}{A}=\frac{y}{B},$$ so the locus is a straight line through the origin.

• When $$A\neq B$$ and $$\delta=\frac{\pi}{2},$$ no simplification of the coefficients makes the two squared terms equal, hence the curve remains a genuine ellipse having unequal semi-axes.

We are now ready to test each option.

Option A: $$A=B$$, $$a=2b$$, $$\delta=\tfrac{\pi}{2}$$. Because the frequencies are in the ratio $$2:1$$, the curve is the famous “figure-of-eight”, not a circle. So Option A is incorrect.

Option B: $$A=B$$, $$a=b$$, $$\delta=\tfrac{\pi}{2}$$. As shown above these conditions give a circle, yet the option claims the curve is a line. Hence Option B is incorrect.

Option C: $$A\neq B$$, $$a=b$$, $$\delta=\tfrac{\pi}{2}$$. Exactly this set of parameters leads to the unequal-axis ellipse discussed earlier. Therefore Option C is correct.

Option D: $$A\neq B$$, $$a=b$$, $$\delta=0$$. These data yield a straight line, whereas the option says parabola, so Option D is also wrong.

Hence, the correct answer is Option C.