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 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?
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Lissajous figures depend on the frequency ratio and phase difference; for equal frequencies and a phase difference of \( \pi/2 \), the motion is elliptical.
Updated On: Jun 9, 2026
- Straight line
- Circular path
- Elliptical path
- Spiral path
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The Correct Option is C
Solution and Explanation
Concept:
The superposition of two simple harmonic motions at right angles to each other, having the same frequency but different amplitudes and phases, is a classic problem in the study of Lissajous figures.
Step 1: Express the motion equations.
The given equations are: $$ x = A \sin(\omega t) \implies \frac{x}{A} = \sin(\omega t) $$ $$ y = B \cos(\omega t) \implies \frac{y}{B} = \cos(\omega t) $$
Step 2: Eliminate time to find the path.
We use the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\): $$ \left( \frac{x}{A} \right)^2 + \left( \frac{y}{B} \right)^2 = \sin^2(\omega t) + \cos^2(\omega t) $$ $$ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 $$
Step 3: Identify the curve.
This is the standard equation of an ellipse centered at the origin. Since \( A \neq B \), the semi-major and semi-minor axes are unequal, resulting in an ellipse. If \( A = B \), the path would reduce to a circular path. $$\boxed{\text{Elliptical path}}$$
Step 1: Express the motion equations.
The given equations are: $$ x = A \sin(\omega t) \implies \frac{x}{A} = \sin(\omega t) $$ $$ y = B \cos(\omega t) \implies \frac{y}{B} = \cos(\omega t) $$
Step 2: Eliminate time to find the path.
We use the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\): $$ \left( \frac{x}{A} \right)^2 + \left( \frac{y}{B} \right)^2 = \sin^2(\omega t) + \cos^2(\omega t) $$ $$ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 $$
Step 3: Identify the curve.
This is the standard equation of an ellipse centered at the origin. Since \( A \neq B \), the semi-major and semi-minor axes are unequal, resulting in an ellipse. If \( A = B \), the path would reduce to a circular path. $$\boxed{\text{Elliptical path}}$$
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