SOLUTION(A) Find the frequencies if the pipe is open at both ends.Substitute into whole harmonicsequation, with n = 1.Multiply to find the second and thirdharmonics.This works out to n = 286.88, whichmust be truncated down (n = 287gives a frequency over 2.00 x 104 Hz).The next two harmonics are oddmultiples of the first:(B) How many harmonics lie between 20 Hz and 20000 Hz for this pipe?Set the frequency in the harmonicsequation equal to 2.00 x 104 Hz andsolve for n.(C) Find the frequencies for the pipe closed at one end.Apply the one end open harmonicsequation with n = 1.f2=f3Vf₁ =-2Lf2 2f1= 139 HzVfn === n2Ln = 286Hz343 m/s2(2.46 m)f1HzVHz4L= 69.7 Hz343 m/s4(2.46 m)f33f1 = 209 Hz343 m/s2. (2.46 m)f3 = 3f₁ = 105 HzLEARN MOREQUESTION True or False: The fundamental wavelength of a longer pipe is greater than the fundamentalwavelength of a shorter pipe.True, because the fundamental wavelength is the length of the pipe.False, because the speed of sound is slower in the longer pipe.True. The exact relation depends on whether the pipe is closed at both ends, but longer pipes ofthe same kind have longer fundamental wavelength.True, because the fundamental wavelength is twice the length of the pipe.True, because the fundamental wavelength is half the length of the pipe.PRACTICE ITUse the worked example above to help you solve this problem. A pipe is 2.42 m long.(a) Determine the frequencies of the first three harmonics if the pipe is open at both ends. Take345 m/s as the speed of sound in air.f171.28Hz=2.00 x 104 Hz= 34.9 Hzf5 = 5f1 = 175 Hz0.56XYour response differs significantly from the correct answer. Rework your solution from thebeginning and check each step carefully. HzHz(b) How many harmonic frequencies of this pipe lie in the audible range, from 20 Hz to 20000 Hz?280(c) What are the three lowest possible frequencies if the pipe is closed at on end and open at theother?f1 =f3 =f5 =

Given data- Length of pipe L=2.42 m speed of sound in air v=345 m/s

Use the worked example above to help you solve this problem. A pipe is 2.42 m long.. (a) Determine the frequencies of the first three harmonics if the pipe is open at both ends. Take 345 m/s as the speed of sound in air. ✓ Hz fi = f2 = f3 = 71.28 0.56 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Hz Hz (b) How many harmonic frequencies of this pipe lie in the audible range, from 20 Hz to 20000 Hz? 280 (c) What are the three lowest possible frequencies if the pipe is closed at on end and open at the other? f1 = f3 = f5= Hz Hz Hz

Use the worked example above to help you solve this problem. A pipe is 2.42 m long.. (a) Determine the frequencies of the first three harmonics if the pipe is open at both ends. Take 345 m/s as the speed of sound in air. ✓ Hz fi = f2 = f3 = 71.28 0.56 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Hz Hz (b) How many harmonic frequencies of this pipe lie in the audible range, from 20 Hz to 20000 Hz? 280 (c) What are the three lowest possible frequencies if the pipe is closed at on end and open at the other? f1 = f3 = f5= Hz Hz Hz

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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