Question

Two waves are represented as
\(y_1 = a_1sin(ωt-\frac{2πx}{λ})\) and
\(y_2 = a_2cos(ωt-\frac{2πx}{λ}+\frac{π}{6})\)
The path difference between two waves is

• A. \(\frac{λ}{5}\)
• B. \(\frac{λ}{4}\)
• C. \(\frac{λ}{3}\)
• D. \(\frac{λ}{2}\)

Hint:

Write both waves with a sine, find the phase difference, then convert to path difference.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Convert the cosine wave into a sine wave so that the phases can be compared.

Step 2: Convert
\[ \cos\theta=\sin\left(\theta+\frac\pi2\right) \]
\[ y_2=a_2\sin\left(\omega t-\frac{2\pi x}{\lambda}+\frac\pi6+\frac\pi2\right)=a_2\sin\left(\omega t-\frac{2\pi x}{\lambda}+\frac{2\pi}{3}\right) \]

Step 3: Phase difference
Compared with \(y_1\), the extra phase is \(\phi=\dfrac{2\pi}{3}\).

Step 4: Path difference
\[ \Delta x=\frac\lambda{2\pi}\phi=\frac{\lambda}{2\pi}\cdot\frac{2\pi}{3}=\frac\lambda3 \]
This is option (C).

Final Answer:
The phase difference is 2 pi over 3, which corresponds to a path difference of lambda over 3, option (C). \[ \boxed{\frac{\lambda}{3}} \]