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A wave of frequency 500 Hz is traveling with a speed of 350 m/s.

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Question

A wave of frequency 500 Hz is traveling with a speed of 350 m/s. (a) What is the phase difference between two displacements at a certain point at times 1.0 ms apart? (b) what will be the smallest distance between two points which are 45° out of phase at an instant of time?

Numerical
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Solution

Given: n = 500 Hz, v = 350 m/s

v = n × λ

∴ λ = `350/500` = 0.7 m

(a) The path difference is the distance covered v × t = 350 × 0.001 = 0.35 m at t = 1.0 ms = 0.001 s.

∴ Phase difference = `(2π)/λxx"Path difference"`

= `(2π)/0.7xx0.35` = π rad

(b) Phase difference = 45° = `π/4` rad

∴ Path difference = `λ/(2π)xx"Phase difference"`

= `0.7/(2π)xxπ/4` = 0.0875 m

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Chapter 6: Superposition of Waves - Exercises [Page 156]

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