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EXERCISE (a) What length of pipe open at both ends has a fundamental frequency of 3.79 x 10² Hz? Find the first overtone. 1pipe = m fovertone = (b) If the one end of this pipe is now closed, what is the new fundamental frequency? Find the first overtone. ffundamental = Hz Hz HINTS: GETTING STARTED I'M STUCK! Sovertone = (c) If the pipe is open at one end only, how many harmonics are possible in the normal hearing range from 20 to 20000 Hz? n = Hz

EXERCISE (a) What length of pipe open at both ends has a fundamental frequency of 3.79 x 10² Hz? Find the first overtone. 1pipe = m fovertone = (b) If the one end of this pipe is now closed, what is the new fundamental frequency? Find the first overtone. ffundamental = Hz Hz HINTS: GETTING STARTED I'M STUCK! Sovertone = (c) If the pipe is open at one end only, how many harmonics are possible in the normal hearing range from 20 to 20000 Hz? n = Hz

College Physics
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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EXERCISE (a) What length of pipe open at both ends has a fundamental frequency of 3.79 x 10² Hz? Find the first overtone. pipe = fovertone m fovertone Hz (b) If the one end of this pipe is now closed, what is the new fundamental frequency? Find the first overtone. ffundamental HINTS: GETTING STARTED I I'M STUCK! Hz Hz (c) If the pipe is open at one end only, how many harmonics are possible in the normal hearing range from 20 to 20000 Hz? n =
Transcribed Image Text:EXERCISE (a) What length of pipe open at both ends has a fundamental frequency of 3.79 x 10² Hz? Find the first overtone. pipe = fovertone m fovertone Hz (b) If the one end of this pipe is now closed, what is the new fundamental frequency? Find the first overtone. ffundamental HINTS: GETTING STARTED I I'M STUCK! Hz Hz (c) If the pipe is open at one end only, how many harmonics are possible in the normal hearing range from 20 to 20000 Hz? n =
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 the pipe is open at both ends. Take 343 m/s as the speed of sound in air. (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 one end and open at the other? STRATEGY Substitute into the all harmonics (open ended) equation for part (a) and the odd numbered harmonics (one end closed) equation for part (c). All harmonics, n = 1, 2, 3 . . . are available for the pipe open 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 harmonics equation, with n = 1. Multiply to find the second and third harmonics. This works out to n = 286.88, which must be truncated down (n = 287 gives a frequency over 2.00 x 104 Hz). f₁ The next two harmonics are odd multiples of the first: = f₂ = 2f1 fn V 2L (B) How many harmonics lie between 20 Hz and 20000 Hz for this pipe? Set the frequency in the harmonics equation equal to 2.00 x 104 Hz and solve for n. f₁ V 2L n = 286 (C) Find the frequencies for the pipe closed at one end. Apply the one end open harmonics equation with n = 1. - = n V 4L 343 m/s 2(2.46 m) f3 = 3f1 = 139 Hz = 69.7 Hz 343 m/s 4(2.46 m) f3 = 3f₁ = 209 Hz 343 m/s 2. (2.46 m) = 105 Hz = 2.00 × 104 Hz = 34.9 Hz f5 = 5f₁= = 175 Hz
Transcribed Image Text: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 the pipe is open at both ends. Take 343 m/s as the speed of sound in air. (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 one end and open at the other? STRATEGY Substitute into the all harmonics (open ended) equation for part (a) and the odd numbered harmonics (one end closed) equation for part (c). All harmonics, n = 1, 2, 3 . . . are available for the pipe open 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 harmonics equation, with n = 1. Multiply to find the second and third harmonics. This works out to n = 286.88, which must be truncated down (n = 287 gives a frequency over 2.00 x 104 Hz). f₁ The next two harmonics are odd multiples of the first: = f₂ = 2f1 fn V 2L (B) How many harmonics lie between 20 Hz and 20000 Hz for this pipe? Set the frequency in the harmonics equation equal to 2.00 x 104 Hz and solve for n. f₁ V 2L n = 286 (C) Find the frequencies for the pipe closed at one end. Apply the one end open harmonics equation with n = 1. - = n V 4L 343 m/s 2(2.46 m) f3 = 3f1 = 139 Hz = 69.7 Hz 343 m/s 4(2.46 m) f3 = 3f₁ = 209 Hz 343 m/s 2. (2.46 m) = 105 Hz = 2.00 × 104 Hz = 34.9 Hz f5 = 5f₁= = 175 Hz
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