Lesson_5.4_Homework_Solution

# Lesson_5.4_Homework_Solution - Lesson 5.4 Homework...

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Unformatted text preview: Lesson 5.4 Homework Solutions 1. p o = ρϖ vs o 3 8 3 10 1.29 2 100 1.1 1 3 0 4 o o m p s v ρϖ π-- × = = = × × × × 2. v 20 = v o + 0.6 t = 332 + 0.6 x 20 = 344 ms-1 . 3. (i) 2 6 , 3 .0 m k π π λ λ = = = (ii) ϖ = 1700 π = 2 π f, f = 850 Hz (iii) v = f λ = 850 x 6 = 5100 ms-1 . (iv) 3 14 0.0025 3.2 10 2.87 17 1 00 00 51 o o p s v m ρϖ π- = = = × × × × 4. (i) 6 12 7.5 10 10lo 68. g 10log 8 1 o d I I B β-- × = = = (ii) 8.5 2 12 35 10log 10 3.5 log 12 lo g 8 1 .5 I I I I W m--- = =- = ⋅ + = 5. The displacement amplitude is given by o o p s v ρϖ = If p 1 is the pressure amplitude of the first wave, p 2 the pressure amplitude of the second wave, ϖ 1 the angular frequency of the first wave and ϖ 2 the angular frequency of the second wave, then we have: 1 2 1 2 1 1 1 2 2 1 2 1 2 p p v v p p ρϖ ρϖ ϖ ϖ ϖ ϖ = = = = The pressure amplitude of the second wave is twice the pressure amplitude of the first wave. Intensity of a wave is directly proportional to the square of its...
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Lesson_5.4_Homework_Solution - Lesson 5.4 Homework...

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