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Question

A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is 350ms1. (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?


Solution

Here given, n =100, v =350 m/s 

λ=(vn)=350100=3.5m

In 2.5 ms, the distance travelled by the particle is given by

Dx=(350×2.5×103)

So, phase difference,

ϕ=2πλ×Dx

=(2π×350×2.5×1033.5)

=(π2)

(b)In the second case,

Given Δη=10cm=101m

So, ϕ=2πxΔx

=2π×101(350100)=2π35

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