Question

Two waves having same amplitude \(A\), angular frequency \(\omega\) and propagation constant \(k\), traveling in opposite directions, have a phase difference of \(60^{\circ }\). If they interfere with each other, the resultant displacement \(y\) of the wave is

• A. \(2A\sin (\omega t+30^{\circ })\cos (kx+30^{\circ })\)
• B. \(A\sin (\omega t+30^{\circ })\cos (kx+30^{\circ })\)
• C. \(2A\sin (kt+30^{\circ })\cos (\omega x+30^{\circ })\)
• D. \(2A\sin (\omega t+kx+60^{\circ })\)

Hint:

Add the two waves using the sum-to-product identity for sines.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Two waves of the same amplitude, frequency and wave number travelling in opposite directions form a stationary wave.

Step 2: Write the waves
\(y_1=A\sin(\omega t-kx)\) and \(y_2=A\sin(\omega t+kx+60^{\circ})\).

Step 3: Add them
Use \(\sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\).
\(\dfrac{C+D}{2}=\omega t+30^{\circ}\) and \(\dfrac{C-D}{2}=-(kx+30^{\circ})\). The cosine is even, so the sign does not matter.
\(y=2A\sin(\omega t+30^{\circ})\cos(kx+30^{\circ})\).

Final Answer:
The resultant is \(2A\sin(\omega t+30^{\circ})\cos(kx+30^{\circ})\). \[ \boxed{\text{(A) }2A\sin (\omega t+30^{\circ })\cos (kx+30^{\circ })} \]