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.f₁ =Hzf₂ =Hzf3Hz=(b) How many harmonic frequencies of this pipe lie in the audible range, from 20 Hz to 20000 Hz?(c) What are the three lowest possible frequencies if the pipe is closed at on end and open at theother?f₁ =f3==f5=HzHzHz GOAL Find frequencies of open and closed pipes.PROBLEM A pipe is 2.46 m long. (a) Determine the frequencies of the first three harmonics if thepipe is open at both ends. Take 343 m/s as the speed of sound in air. (b) How many harmonicfrequencies of this pipe lie in the audible range, from 20 Hz to 20000 Hz? (c) What are the three lowestpossible frequencies if the pipe is closed at one end and open at the other?STRATEGY Substitute into the all harmonics (open ended) equation for part (a) and the odd numberedharmonics (one end closed) equation for part (c). All harmonics, n = 1, 2, 3 . . . are available for the pipeopen at both ends, but only the harmonics with n = 1, 3, 5, . . . for the pipe closed at one end. For part(b), set the frequency in the odd numbered harmonics equation equal to 2.00 x 104 Hz.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).f₁The next two harmonics are oddmultiples of the first:=f₂ = 2f1fnV2L(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.f₁V2Ln = 286(C) Find the frequencies for the pipe closed at one end.Apply the one end open harmonicsequation with n = 1.-= nV4L343 m/s2(2.46 m)f3 = 3f1=139 Hz= 69.7 Hz343 m/s4(2.46 m)f3 = 3f₁ = 209 Hz343 m/s2. (2.46 m)= 105 Hz= 2.00 × 104 Hz= 34.9 Hzf5 = 5f₁== 175 Hz

PRACTICE IT : Length of pipe (L) = 2.42 m

PRACTICE IT 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. f₁ = Hz f₂ = Hz √3 Hz (b) How many harmonic frequencies of this pipe lie in the audible range, from 20 Hz to 20000 Hz? (c) What are the three lowest possible frequencies if the pipe is closed at on end and open at the other? f₁= √3= fs= Hz Hz Hz

PRACTICE IT 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. f₁ = Hz f₂ = Hz √3 Hz (b) How many harmonic frequencies of this pipe lie in the audible range, from 20 Hz to 20000 Hz? (c) What are the three lowest possible frequencies if the pipe is closed at on end and open at the other? f₁= √3= fs= 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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