Question

The equations of two waves acting in perpendicular directions are given as$x=a \cos(\omega t+\delta)$ and $y=a \cos(\omega t+\alpha),$ where $\delta=\alpha+\frac{\pi}{2}, $ the resultant wave represents

• A. a circle (c.w)
• B. a circle (a.c.w)
• C. an Ellipse (c.w)
• D. an ellipse (a.c.w)

Answer 1

The Correct Option is B

To determine the nature of the resultant wave produced by two perpendicular waves, we start by analyzing the given equations:

• First wave: x = a \cos(\omega t + \delta)
• Second wave: y = a \cos(\omega t + \alpha)
• Phase relationship: \delta = \alpha + \frac{\pi}{2}

This indicates the phase difference between the waves is \frac{\pi}{2} or 90 degrees.

Analysis and Explanation:

Given the phase relationship, we can substitute \delta in the equation for x:

x = a \cos\left(\omega t + \alpha + \frac{\pi}{2}\right)

We know the trigonometric identity:

\cos\left(\theta + \frac{\pi}{2}\right) = -\sin(\theta)

Applying this, the equation for x becomes:

x = a(-\sin(\omega t + \alpha)) or x = -a \sin(\omega t + \alpha)

Now we have:

• x = -a \sin(\omega t + \alpha)
• y = a \cos(\omega t + \alpha)

To find the nature of the locus of (x, y) as t varies, we use the following trigonometric identity:

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

Apply this identity:

\left(\frac{x}{a}\right)^2 + \left(\frac{y}{a}\right)^2 = \sin^2(\omega t + \alpha) + \cos^2(\omega t + \alpha) = 1

Simplifying gives:

x^2 + y^2 = a^2

This equation represents a circle. The direction of traversal can be determined by the negative sign in the expression for x, which indicates anticlockwise (a.c.w) rotation.

Conclusion:

The resultant wave represents a circle traced in the anticlockwise direction (a.c.w).