Question

Two sound waves each of wavelength \(λ\) and same amplitude \(A\) interfere at point \(Q\). If the path difference is \(\frac{λ}{4}\), the amplitude of the resultant wave at point \(Q\) is \([sin\frac{π}{2} = 1,cos\frac{π}{2} = 0]\)

• A. \(A\)
• B. \(\sqrt{2}A\)
• C. \(3A\)
• D. \(\sqrt{3}A\)

Hint:

A path difference of lambda over 4 means a phase difference of pi over 2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The phase difference is \(\phi=\dfrac{2\pi}{\lambda}\times\text{path difference}\).

Step 2: Phase difference
\[ \phi=\frac{2\pi}{\lambda}\cdot\frac\lambda4=\frac\pi2 \]

Step 3: Resultant amplitude
For two waves of equal amplitude \(A\):
\[ A_R=2A\cos\frac\phi2=2A\cos\frac\pi4=2A\cdot\frac1{\sqrt2}=\sqrt2A \]

Step 4: Check the options
Amplitude \(A\) would need a phase difference of \(120^\circ\). Amplitude \(3A\) is impossible because the maximum is \(2A\). So option (B).

Final Answer:
The phase difference is pi/2, so the amplitude is root 2 times A, option (B). \[ \boxed{\sqrt2A} \]