Explanation Given info: The length of the string is 0.400 m , the mass per unit length is 9.00 × 10 − 3 kg / m . The length of the pipe opened at one end in the condition of second resonance is, L p = 3 λ 4 λ = 4 L p 3 . (1) Here, λ is the wavelength. L p is the length of pipe. Substitute 1.75 m for L p in equation (1). λ = 4 × 1.75 m 3 = 2.33 m The formula for the frequency of the string is, f = v λ Here, v is the speed of sound. Substitute 343 m / s for v and 2.33 m for λ in above equation. f = 343 m / s 2.33 m = 147.21 Hz The length of string is equal to the wavelength. λ = L s Substitute 0

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN: 9781305116399

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Chapter 18, Problem 18.63AP

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The tension in string if its second harmonic has the same frequency as the second resonance mode of a 1.75 m long pipe open at one end.

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Chapter 18, Problem 18.62AP

Chapter 18, Problem 18.64AP

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Physics for Scientists and Engineers, Technology Update (No access codes included)

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The tension in string if its second harmonic has the same frequency as the second resonance mode of a 1.75 m long pipe open at one end.

Solution Summary: The author determines the tension in string if its second harmonic has the same frequency as the second resonance mode of a 1.75m long pipe.

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The tension in string if its second harmonic has the same frequency as the second resonance mode of a 1.75 m long pipe open at one end.

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Chapter 18 Solutions