Question

A progressive wave is given by \(Y=6\,\sin (3t-2x)\) m. On this wave, how far away are the two points having phase difference of \(45^{\circ }\) ?

• A. \(\frac{\pi }{16}\text{m}\)
• B. \(\frac{\pi }{8}\text{m}\)
• C. \(\frac{\pi }{4}\text{m}\)
• D. \(\frac{\pi }{2}\text{m}\)

Hint:

Phase difference equals wave number times separation; get the wave number from the equation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Compare \(Y=6\sin(3t-2x)\) with \(Y=A\sin(\omega t-kx)\): \(k=2\ \text{rad/m}\).

Step 2: Calculation
\(\Delta\phi=k\,\Delta x\) with \(\Delta\phi=45^{\circ}=\dfrac{\pi}{4}\).

Step 3: Solve
\(\Delta x=\dfrac{\pi/4}{2}=\dfrac{\pi}{8}\ \text{m}\)

Final Answer:
The two points are pi over 8 metres apart. \[ \boxed{\text{(B) }\dfrac{\pi}{8}\text{m}} \]