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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Only need to solve part c
Suppose you shake a slinky continuously in an up and down fashion such that a transverse waves propagate horizontally down its length. If you count 15 peaks have been made during a time of 2.03s, what is the frequency of the wave (in Hz)? V Note: In the space below, please enter you numerical answer. Do not enter any units. If you enter units, your answer will be marked as incorrect.
SOLUTION (A) Find the first three harmonics at the given tension. Calculate the speed of the wave on the wire: Find the wire's fundamental frequency: Find the next two harmonics by multiplication: = V = f₁ Substitute the values of F, μ, and L. F 1/2 80 N (=) ¹/² = (2.00 x 10-³ kg/m = (B) Find the wavelength of the sound waves produced. Solve vs fλ for the wavelength and substitute the frequencies. F == A V f₁ 2L (C) Find the fundamental frequency corresponding to the elastic limit. Calculate the tension in the wire from the elastic limit: f₁ f₂ = 2f₁ = 2.00 × 10² Hz, ƒ3 = 3ƒ₁ = 3.00 × 10² Hz A₁ = v₁/f₁ = (345 m/s)/(1.00 x 10² Hz) = 3.45 m 1₂ = vs/f₂ = (345 m/s)/(2.00 x 10² Hz) = 1.73 m ^₂ = vs/f3 = (345 m/s)/(3.00 x 10² Hz) = 1.15 m = 2.00 × 10² m/s 2(1.00 m) elastic limit →→ F = (elastic limit) A F = (2.80 x 108 Pa) (2.56 × 10-7m²) = 71.7 N 1 V 2L F 1/2 -= 1.00 x 10² Hz μ = 2.00 x 10² m/s 1 2(1.00 m) 71.7 N 2.00 x 10-³ kg/m = 94.7 Hz
A 1.1 m string oscillating at a frequency of 86 Hz generates the standing wave shown below. What is the wave speed on the string. What is the linear density on the string if the string tension is 5.6 N?
Needs Complete solution with 100 % accuracy.
Please provide details about your resolution process. Please show complete, organized and correct work for each question. Questions without complete and clear work will earn no points. You’re welcome to use your textbook and notes, but no calculators. One way to tune a piano is to strike a tuning fork (which emits only one specific frequency), then immediately strike the piano key for the frequency being sounded by the fork, and listen for beats. If the frequency of an out-of-tune A4string is 420.0 Hz, and a tuner strikes an A4 tuning fork (frequency 440.0 Hz),a. …what beat frequency will the tuner hear?a) ????? = _____________________b. …and what frequency note will the tuner hear?b) ????? = _____________________