A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is 350 m s-1. (a) By how much is the phase changed at a given point in 2.5 ms? (b) What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation?

Question

Navjyot Kalra · Answer

Sol. a) Here given n = 100, v = 350 m/s ⇒ λ = v/m = 350/100 = 3.5 In 2.5 ms, the distance travelled by the particle is given by ∆* = 350 * 2.5 * 10^-3 So, phase difference ∅ = 2π/λ * ∆* ⇒ 2π/(350/100) * 350 * 2.5 * 10^-3 = (π/2). b) In the second case, Given ∆η = 10 cm = 10^-1 m. So, ϕ = 2π/*∆* = 2π * 10^-1/(350/100) = 2π/35.

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