Question

Two simple harmonic progressive waves have displacements \(y_{1}=5\,\sin \left(\frac{2\pi x}{\lambda }-\omega t\right)\) and \(y_{2}=5\,\cos \left(\frac{2\pi x}{\lambda }-\omega t+\frac{\pi }{6}\right)\). The phase difference between the two waves is

• A. \(\frac{\pi }{3}\)
• B. \(\frac{\pi }{6}\)
• C. \(\frac{2\pi }{3}\)
• D. \(\frac{3\pi }{2}\)

Hint:

Rewrite the cosine wave as a sine wave and compare the phases.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Let \(\theta=\dfrac{2\pi x}{\lambda}-\omega t\). The first wave is \(y_1=5\sin\theta\).

Step 2: Convert cosine to sine
\(y_2=5\cos\left(\theta+\dfrac{\pi}{6}\right)=5\sin\left(\theta+\dfrac{\pi}{6}+\dfrac{\pi}{2}\right)=5\sin\left(\theta+\dfrac{2\pi}{3}\right)\)

Step 3: Phase difference
\(\Delta\phi=\dfrac{2\pi}{3}-0=\dfrac{2\pi}{3}\)

Final Answer:
The phase difference is two pi by three. \[ \boxed{\text{(C) }\dfrac{2\pi}{3}} \]