Question

Motion of a particle in x–y plane is described by a set of following
\(x = 4sin (\frac{π}{2} - ωt )m\)
and y = 4sin(ωt) m. The path of the particle will be

• A. Circular
• B. Helical
• C. Parabolic
• D. Elliptical

Answer 1

The Correct Option is A

To determine the path of the particle described by the given equations in the \(x\)–\(y\) plane, we first need to analyze the equations:

Given:

x = 4 \sin \left(\frac{\pi}{2} - \omega t\right) \, \text{m}

y = 4 \sin(\omega t) \, \text{m}

We start by simplifying the equation for \(x\):

Using the trigonometric identity, \(\sin\left(\frac{\pi}{2} - \theta\right) = \cos(\theta)\), we can rewrite the expression for \(x\):

x = 4 \cos(\omega t)

Now, we have:

x = 4 \cos(\omega t)

y = 4 \sin(\omega t)

Next, we need to determine the relationship between \(x\) and \(y\). To do this, let's use the known trigonometric identity:

\cos^2(\theta) + \sin^2(\theta) = 1

Substitute \(x\) and \(y\) into this identity:

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

Since \(\cos^2(\omega t) + \sin^2(\omega t) = 1\), we have:

\frac{x^2}{16} + \frac{y^2}{16} = 1

Multiplying through by 16 gives:

x^2 + y^2 = 16

This equation is in the standard form of a circle, which is x^2 + y^2 = r^2, where \(r\) is the radius of the circle. Here, \(r^2 = 16\) implies \(r = 4\).

Therefore, the path of the particle is a circle with radius 4.

Conclusion: The path of the particle is circular, hence the correct answer is "Circular".