Question

A sound wave of frequency 100 Hz is traveling in air. The speed of sound in air is 350 m/s.
(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 10.0 cm along the direction of propagation?

Answer 1

Answer ⇒ π/2 and 2π/35.

Explanation ⇒  

(a). The wavelength of the sound wave = speed/frequency

∴ λ = 350/100 m

∴ λ = 3.5 m

Now, Time (t) = 2.5 ms

= 2.5/1000 s.

= 0.0025 s .

The small distance traveled in this time,

Δx = Vt

= 350 × 0.0025 m

= 0.875 m.

Hence the phase difference  = 2π × Δx/λ

= 2π × 0.875/3.5

= 0.5 × π = π/2

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(b). Here Δx = 10.0 cm = 0.10 m

Hence the phase difference Ф = 2π × Δx/λ

= 2π × 0.10/3.5

= 2π/35

Hope it helps.

Answer 2

here is ur answer hpe it helps uh