Question

If the resultant of two waves having amplitude \(\mathrm{b}\) is \(\mathrm{b}\), then the phase difference between the two waves is (A) \(120^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(180^{\circ}\)

Short answer

The phase difference between the two waves is \(90^{\circ}\).

Step-by-step solution

Write down the superposition equation for two waves with amplitude b and phase difference φ

The superposition equation for the resultant of two waves with amplitude b and a phase difference of φ is: \[R = \sqrt{b^2 + b^2 + 2bb \cos(\phi)}\]

Solve the equation for R=b

Since we know that R = b, we can substitute it in the equation and solve for φ to find the phase difference. \[b = \sqrt{b^2 + b^2 + 2bb \cos(\phi)}\]

Solve for the phase difference φ

We will now simplify the equation and solve for φ: \[b^2 = b^2 + b^2 + 2b^2 \cos(\phi)\] \[0 = b^2 + 2b^2 \cos(\phi)\] Using the property for cos(θ) : \[\cos(90^{\circ}- \phi)=-\cos(\phi)\] Divide by b^2 on each side of the equation: \[0 = 1 + 2 \cos(\phi)\] Now, substitute the new property of cos(θ) in the equation, \[0 = 1 - 2\cos(90^{\circ} - \phi)\]

Discard the possible multiple solutions

The above equation can have multiple solutions for the value of φ, as the equation depends on cos(θ). However, it is important to consider that the phase difference must be within the range from 0° to 180°.

Find the correct option

Comparing our equation with the given options, we find a match with the option (C): (C) \(90^{\circ}\) Therefore, the phase difference between the two waves is \(90^{\circ}\).