Question

Two simple harmonic motions, as shown below, 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.

• A. Parameters \(A\ne B,\ a=b;\ \delta=0\);
  Curve Parabola
• B. Parameters \(A=B,\ a=b;\ \delta=\frac{\pi}{2}\)
  Curve Line
• C. Parameters \(A\ne B,\ a=b;\ \delta=\frac{\pi}{2}\)
  Curve Ellipse
• D. Parameters \(A=B,\ a=2b;\ \delta=\frac{\pi}{2}\)
  Curve Circle

Hint:

Eliminate t in each case: equal frequencies with phase \(\pi/2\) give an ellipse (a circle only if the amplitudes are also equal); \(\delta=0\) gives a straight line.

Answer 1

The Correct Option is C

Option 1: a = b, \(\delta=0\): \(\frac{x}{A}=\sin at,\ \frac{y}{B}=\sin at\), so \(y=\frac{B}{A}x\), a straight line, not a parabola. Wrong.
Option 2: A = B, a = b, \(\delta=\frac{\pi}{2}\): \(x=A\cos at,\ y=A\sin at\), so \(x^2+y^2=A^2\), a circle, not a line. Wrong.
Option 3: a = b, \(\delta=\frac{\pi}{2}\), A not equal to B: \(\frac{x^2}{A^2}+\frac{y^2}{B^2}=1\), an ellipse. Correct.
Option 4: a = 2b, \(\delta=\frac{\pi}{2}\), A = B: \(x=A\cos2bt=A(1-2\sin^2bt)\), \(y=A\sin bt\), so \(x=A-\frac{2y^2}{A}\), a parabola, not a circle. Wrong.
Answer: (3) Parameters \(A\ne B,\ a=b;\ \delta=\frac{\pi}{2}\); Curve Ellipse