Question

Consider the superposition of two orthogonal simple harmonic motions \(y_1 = a \cos 2\omega t\) and \(y_2 = b \cos(\omega t + \phi)\). If \(\phi = \pi\), the resultant motion will represent
(a, b and \(\omega\) are constants with appropriate dimensions)

• A. a parabola
• B. a hyperbola
• C. an ellipse
• D. a circle

Hint:

Lissajous figures are determined by the frequency ratio and phase difference. When the frequency ratio is 1:2 (or 2:1), the resulting curve is often a parabola or a figure-eight shape. The key is to use the double-angle trigonometric identity (\(\cos(2\theta) = 2\cos^2\theta - 1\)) to eliminate the time variable.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem describes the superposition of two perpendicular simple harmonic motions (SHMs) with different frequencies. The path traced by the particle under this combined motion is known as a Lissajous figure. To determine the shape of the path, we need to eliminate the time variable (\(t\)) from the given parametric equations. Let's assign the motions to the x and y axes, for example, \(x(t) = y_2\) and \(y(t) = y_1\).
Step 2: Key Formula or Approach:
The parametric equations of the motion are: \[ x(t) = b \cos(\omega t + \phi) \] \[ y(t) = a \cos(2\omega t) \] We are given the phase \(\phi = \pi\). We will use trigonometric identities to eliminate \(t\). The relevant identity is the double-angle formula: \(\cos(2\theta) = 2\cos^2(\theta) - 1\).
Step 3: Detailed Explanation:
1. Write down the parametric equations with the given phase: Let the orthogonal motions be along the x and y axes. \[ x = b \cos(\omega t + \pi) \] \[ y = a \cos(2\omega t) \] 2. Simplify the equation for x: Using the identity \(\cos(\theta + \pi) = -\cos(\theta)\), the x-equation becomes: \[ x = -b \cos(\omega t) \] 3. Eliminate the time variable \(t\): From the simplified x-equation, we can express \(\cos(\omega t)\) in terms of \(x\): \[ \cos(\omega t) = -\frac{x}{b} \] Now, use the double-angle identity for the y-equation: \[ y = a \cos(2\omega t) = a (2\cos^2(\omega t) - 1) \] Substitute the expression for \(\cos(\omega t)\) into the y-equation: \[ y = a \left[ 2 \left(-\frac{x}{b}\right)^2 - 1 \right] \] \[ y = a \left( \frac{2x^2}{b^2} - 1 \right) \] 4. Identify the shape of the curve: Rearranging the equation, we get: \[ y = \left(\frac{2a}{b^2}\right)x^2 - a \] This equation is of the form \(y = kx^2 + c\), where \(k = \frac{2a}{b^2}\) and \(c = -a\) are constants. This is the standard equation of a parabola that opens vertically. Step 4: Final Answer:
The equation relating y and x is \(y = (\frac{2a}{b^2})x^2 - a\), which represents a parabola. Therefore, option (A) is correct.