Question

A sound wave is travelling with a frequency of $50\ \text{Hz}$. The phase difference between the two points in the path of a wave is $\frac{\pi}{3}$. The distance between those two points is (Velocity of sound in air $= 330\ \text{m/s}$)

• A. $1.1\ \text{m}$
• B. $0.6\ \text{m}$
• C. $2.2\ \text{m}$
• D. $1.7\ \text{m}$

Hint:

Think of phase difference as a fractional portion of a full wave cycle. A full cycle of $2\pi$ radians corresponds to exactly one full wavelength ($\lambda = 6.6\ \text{m}$). The given phase difference is $\frac{\pi}{3}$, which is exactly $\frac{1}{6}\text{th}$ of a full $2\pi$ cycle ($\frac{\pi/3}{2\pi} = \frac{1}{6}$). Therefore, the distance must simply be $\frac{1}{6}\text{th}$ of the wavelength: $\frac{6.6}{6} = 1.1\ \text{m}$. This mental shortcut completely bypasses formal algebraic rearrangements!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A sound wave of frequency 50 Hz travels at 330 m/s. Two points have a phase difference Δφ = π/3. Find the physical distance Δx between them.

Step 2: Key Formula or Approach:
Wavelength λ = v/f. Phase difference relates to path difference by Δφ = (2π/λ)·Δx, so Δx = (λ/2π)·Δφ.

Step 3: Detailed Explanation:
Compute λ = 330/50 = 6.6 m. Substitute into the path difference formula: Δx = (6.6/2π) × (π/3). The π cancels, leaving Δx = 6.6/(2×3) = 6.6/6 = 1.1 m.

Step 4: Final Answer:
The distance between the points is 1.1 m, option (A).