Question

The path difference between the two waves $y_{1} = a_{1}\sin \left(\omega t - \frac{2\pi x}{\lambda}\right)$ and $y_{2} = a_{2}\cos \left(\omega t - \frac{2\pi x}{\lambda} +\phi\right)$ is

• A. $\frac{\lambda}{2\pi}\phi$
• B. $\frac{\lambda}{2\pi}\left(\phi +\frac{\pi}{2}\right)$
• C. $\frac{2\pi}{\lambda}\left(\phi -\frac{\pi}{2}\right)$
• D. $\frac{2\pi}{\lambda}\phi$

Hint:

Convert cosine to sine: $\cos\theta = \sin\left(\theta + \frac{\pi}{2}\right)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Convert cosine to sine: $\cos\theta = \sin\left(\theta + \frac{\pi}{2}\right)$.
Step 2: Detailed Explanation:
Write $y_2 = a_2 \sin\left(\omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2}\right)$. Phase difference $\Delta\phi = \phi + \frac{\pi}{2}$. Path difference $\Delta x = \frac{\lambda}{2\pi} \Delta\phi = \frac{\lambda}{2\pi}\left(\phi + \frac{\pi}{2}\right)$.
Step 3: Final Answer:
The path difference is $\frac{\lambda}{2\pi}\left(\phi +\frac{\pi}{2}\right)$.