Question

A particle is subjected to two simple harmonic motions along the x and y axes, described by x(t) = \(\alpha\)sin(2\(\omega\)t + π) and y(t) = 2\(\alpha\) sin(\(\omega\)t). The resultant motion is given by

• A. \(\frac{x^2}{\alpha^2}+\frac{y^2}{4\alpha^2}=1\)
• B. x²+y²=1
• C. \(y^2=x^2(1-\frac{x^2}{4\alpha^2})\)
• D. \(x^2=y^2(1-\frac{y^2}{4\alpha^2})\)

Answer 1

The Correct Option is D

To solve the problem, let's analyze the given simple harmonic motions along the x and y axes:

For the x-axis motion:

\(x(t) = \alpha \sin(2\omega t + \pi)\)

This can be simplified using the trigonometric identity \(\sin(\theta + \pi) = -\sin(\theta)\):

\(x(t) = -\alpha \sin(2\omega t)\)

For the y-axis motion:

\(y(t) = 2\alpha \sin(\omega t)\)

We will use the parametric equations for the x and y motions to find the resultant equation of motion. Combining both equations:

Squaring both equations and adding:

• \(x^2 = \alpha^2 \sin^2(2\omega t)\)
• \(y^2 = 4\alpha^2 \sin^2(\omega t)\)

Recall the double angle identity: \(\sin(2\omega t) = 2\sin(\omega t)\cos(\omega t)\).

Substituting this into x²:

\(x^2 = \alpha^2 (2\sin(\omega t)\cos(\omega t))^2 = 4\alpha^2 \sin^2(\omega t) \cos^2(\omega t)\)

Thus, we have:

• \(x^2 = 4\alpha^2 \sin^2(\omega t) \cos^2(\omega t)\)

Substitute \(\sin^2(\omega t) = \frac{y^2}{4\alpha^2}\) into the equation for x²:

\(x^2 = 4\alpha^2 \left( \frac{y^2}{4\alpha^2} \right) \cos^2(\omega t)\)

\(x^2 = y^2 \cos^2(\omega t)\)

Recognizing that:

\(\cos^2(\omega t) = 1 - \sin^2(\omega t)\)

Leads to:

\(x^2 = y^2 \left( 1 - \frac{y^2}{4\alpha^2} \right)\)

This corresponds to the resultant motion \(x^2 = y^2 \left(1 - \frac{y^2}{4\alpha^2}\right)\), which matches option 4. This is the path described by the particle, which typically represents an ellipse.