Question

\(y_1 = 4\sin(\Omega t + kx)\), \(y_2 = -4\cos(\Omega t + kx)\). The phase difference is

• A. \(\pi/2\)
• B. \(3\pi/2\)
• C. \(\pi\)
• D. zero

Hint:

\(-\cos\theta = \sin(\theta - \pi/2) = \sin(\theta + 3\pi/2)\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Convert to same trigonometric function.
Step 2: Detailed Explanation:
\(y_2 = -4\cos(\Omega t + kx) = 4\sin(\Omega t + kx - \pi/2)\)
Also = \(4\sin(\Omega t + kx + 3\pi/2)\)
Phase difference = \(3\pi/2\)
Step 3: Final Answer:
Phase difference is \(3\pi/2\).