Question

A sound wave of frequency $160\text{ Hz}$ has a velocity of $320\text{ m/s}$. When it travels through air, the particles having a phase difference of $90^\circ$ are separated by a distance of

• A. $50\text{ cm}$
• B. $1\text{ cm}$
• C. $25\text{ cm}$
• D. $75\text{ cm}$

Hint:

Think of phase difference as a fraction of a full wave cycle. A phase difference of $90^\circ$ is exactly one-quarter of a full $360^\circ$ cycle. Therefore, the distance separation is simply one-quarter of the wavelength: $\Delta x = \frac{\lambda}{4} = \frac{200\text{ cm}}{4} = 50\text{ cm}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Sound wave with f = 160 Hz, v = 320 m/s; find path separation Δx between points with phase difference 90°.

Step 2: Key Formula or Approach:
Wavelength λ = v/f. Phase difference Δφ = (2π/λ)·Δx. Convert 90° to π/2 radians.

Step 3: Detailed Explanation:
λ = 320/160 = 2 m = 200 cm. π/2 = (2π/200)·Δx → 1/2 = Δx/100 → Δx = 50 cm.

Step 4: Final Answer:
Path separation is 50 cm, matching option (A).