Question

A particle executes two simple harmonic motions along mutually perpendicular axes, given by \( x = A \sin(\omega_1 t) \) and \( y = B \cos(\omega_2 t) \) where \( A \neq B \) and \( \omega_1 = \omega_2 \). Which of the following best describes the resultant motion of the particle?

• A. Straight line
• B. Circular path
• C. Elliptical path
• D. Spiral path

Hint:

Lissajous figures depend on the frequency ratio and phase difference; for equal frequencies and a phase difference of \( \pi/2 \), the motion is elliptical.

Answer 1

The Correct Option is C

Step 1: Read off the two motions.
The particle moves along $x$ as $x = A\sin(\omega t)$ and along $y$ as $y = B\cos(\omega t)$, with the same frequency $\omega_1=\omega_2=\omega$. Both coordinates oscillate in time, so the particle traces some closed curve in the $x$-$y$ plane.
Step 2: Make a clean fraction for each coordinate.
Divide each equation by its amplitude so the trig functions stand alone: \[ \frac{x}{A} = \sin(\omega t), \qquad \frac{y}{B} = \cos(\omega t). \] This is the key trick, it isolates $\sin$ and $\cos$ of the same angle.
Step 3: Remove time using one identity.
Square each and add, because $\sin^2\theta + \cos^2\theta = 1$ wipes out the time: \[ \left(\frac{x}{A}\right)^2 + \left(\frac{y}{B}\right)^2 = \sin^2(\omega t) + \cos^2(\omega t) = 1. \]
Step 4: Recognise the standard shape.
So the path satisfies $\dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} = 1$, which is exactly the equation of an ellipse centred at the origin with axes along $x$ and $y$.
Step 5: Use the condition $A \neq B$.
If $A$ and $B$ were equal the two denominators would match and the curve would be a circle. Here we are told $A \neq B$, so the horizontal and vertical extents differ and the curve stays a genuine ellipse, not a circle.
Step 6: State the resultant motion.
There is no growing radius (nothing spirals) and it is not a single straight line, since neither coordinate is a fixed multiple of the other. The motion is therefore an elliptical path.
\[ \boxed{\text{Elliptical path}} \]